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class | __index_level_0__ int64 0 742k |
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## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0}\left(2-3^{\operatorname{arctg}^{2} \sqrt{x}}\right)^{\frac{2}{\sin x}}$ | ## Solution
$$
\begin{aligned}
& \lim _{x \rightarrow 0}\left(2-3^{\operatorname{arctg}^{2} \sqrt{x}}\right)^{\frac{2}{\sin x}}= \\
& =\lim _{x \rightarrow 0}\left(e^{\ln \left(2-3^{\operatorname{arctg}^{2} \sqrt{x}}\right)}\right)^{\frac{2}{\sin x}}= \\
& =\lim _{x \rightarrow 0} e^{\frac{2 \ln \left(2-3^{\operatorna... | \frac{1}{9} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,893 |
## Problem Statement
Calculate the limit of the function:
$$
\lim _{x \rightarrow 0}\left(\frac{e^{3 x}-1}{x}\right)^{\cos ^{2}\left(\frac{\pi}{4}+x\right)}
$$ | ## Solution
$$
\begin{aligned}
& \lim _{x \rightarrow 0}\left(\frac{e^{3 x}-1}{x}\right)^{\cos ^{2}\left(\frac{\pi}{4}+x\right)}=\left(\lim _{x \rightarrow 0} \frac{e^{3 x}-1}{x}\right)^{\lim _{x \rightarrow 0} \cos ^{2}\left(\frac{\pi}{4}+x\right)}= \\
& =\left(\lim _{x \rightarrow 0} \frac{e^{3 x}-1}{x}\right)^{\cos... | \sqrt{3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,894 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 2}\left(\frac{\cos x}{\cos 2}\right)^{\frac{1}{x-2}}$ | ## Solution
$\lim _{x \rightarrow 2}\left(\frac{\cos x}{\cos 2}\right)^{\frac{1}{x-2}}=\lim _{x \rightarrow 2}\left(e^{\ln \left(\frac{\cos x}{\cos 2}\right)}\right)^{\frac{1}{x-2}}=$
$=\lim _{x \rightarrow 2} e^{\frac{1}{x-2} \cdot \ln \left(\frac{\cos x}{\cos 2}\right)}=\exp \left\{\lim _{x \rightarrow 2} \frac{1}{... | e^{-\tan2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,895 |
## Problem Statement
Calculate the limit of the function:
$$
\lim _{x \rightarrow 2}(\sin x)^{\frac{3}{1+x}}
$$ | ## Solution
$\lim _{x \rightarrow 2}(\sin x)^{\frac{3}{1+x}}=(\sin 2)^{\frac{3}{1+2}}=(\sin 2)^{1}=\sin 2$
## Problem Kuznetsov Limits 20-4 | \sin2 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,896 |
## Problem Statement
$\lim _{x \rightarrow 0} \frac{\tan x \cdot \cos \left(\frac{1}{x}\right)+\log (2+x)}{\log (4+x)}$ | ## Solution
Since $\cos \left(\frac{1}{x}\right)_{\text {- is bounded, and }} \operatorname{tg} x \rightarrow 0$, as $x \rightarrow 0$, then
$$
\operatorname{tg} x \cdot \cos \left(\frac{1}{x}\right) \rightarrow 0, \text { as } x \rightarrow 0
$$
Then:
$\lim _{x \rightarrow 0} \frac{\operatorname{tg} x \cdot \cos \... | \frac{1}{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,897 |
## Problem Statement
Find the indefinite integral:
$$
\int \frac{x^{3}+4 x^{2}+4 x+2}{(x+1)^{2}\left(x^{2}+x+1\right)} d x
$$ | ## Solution
$$
\int \frac{x^{3}+4 x^{2}+4 x+2}{(x+1)^{2}\left(x^{2}+x+1\right)} d x=
$$
We decompose the proper rational fraction into partial fractions using the method of undetermined coefficients:
$$
\begin{aligned}
& \frac{x^{3}+4 x^{2}+4 x+2}{(x+1)^{2}\left(x^{2}+x+1\right)}=\frac{A_{1}}{x+1}+\frac{A_{2}}{(x+1)... | -\frac{1}{x+1}+\frac{1}{2}\cdot\ln(x^{2}+x+1)+\frac{1}{\sqrt{3}}\cdot\operatorname{arctg}(\frac{2x+1}{\sqrt{3}})+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,898 |
## Problem Statement
Find the indefinite integral:
$$
\int \frac{x^{3}+4 x^{2}+3 x+2}{(x+1)^{2}\left(x^{2}+1\right)} d x
$$ | ## Solution
$$
\int \frac{x^{3}+4 x^{2}+3 x+2}{(x+1)^{2}\left(x^{2}+1\right)} d x=
$$
Decompose the proper rational fraction into partial fractions using the method of undetermined coefficients:
$$
\begin{aligned}
& \frac{x^{3}+4 x^{2}+3 x+2}{(x+1)^{2}\left(x^{2}+1\right)}=\frac{A_{1}}{x+1}+\frac{A_{2}}{(x+1)^{2}}+\... | -\frac{1}{x+1}+\frac{1}{2}\ln|x^{2}+1|+\operatorname{arctg}x+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,899 |
## Problem Statement
Find the indefinite integral:
$$
\int \frac{2 x^{3}+7 x^{2}+7 x-1}{(x+2)^{2}\left(x^{2}+x+1\right)} d x
$$ | ## Solution
$$
\int \frac{2 x^{3}+7 x^{2}+7 x-1}{(x+2)^{2}\left(x^{2}+x+1\right)} d x=
$$
We decompose the proper rational fraction into partial fractions using the method of undetermined coefficients:
$$
\begin{aligned}
& \frac{2 x^{3}+7 x^{2}+7 x-1}{(x+2)^{2}\left(x^{2}+x+1\right)}=\frac{A_{1}}{x+2}+\frac{A_{2}}{(... | \frac{1}{x+2}+\ln(x^{2}+x+1)-\frac{2}{\sqrt{3}}\cdot\operatorname{arctg}(\frac{2x+1}{\sqrt{3}})+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,900 |
## Problem Statement
Find the indefinite integral:
$$
\int \frac{2 x^{3}+4 x^{2}+2 x-1}{(x+1)^{2}\left(x^{2}+2 x+2\right)} d x
$$ | ## Solution
$$
\int \frac{2 x^{3}+4 x^{2}+2 x-1}{(x+1)^{2}\left(x^{2}+2 x+2\right)} d x=
$$
We decompose the proper rational fraction into partial fractions using the method of undetermined coefficients:
$$
\begin{aligned}
& \frac{2 x^{3}+4 x^{2}+2 x-1}{(x+1)^{2}\left(x^{2}+2 x+2\right)}=\frac{A_{1}}{x+1}+\frac{A_{2... | \frac{1}{x+1}+\ln(x^{2}+2x+2)-\operatorname{arctg}(x+1)+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,901 |
## Problem Statement
Find the indefinite integral:
$$
\int \frac{x^{3}+6 x^{2}+9 x+6}{(x+1)^{2}\left(x^{2}+2 x+2\right)} d x
$$ | ## Solution
$$
\int \frac{x^{3}+6 x^{2}+9 x+6}{(x+1)^{2}\left(x^{2}+2 x+2\right)} d x=
$$
We decompose the proper rational fraction into partial fractions using the method of undetermined coefficients:
$$
\begin{aligned}
& \frac{x^{3}+6 x^{2}+9 x+6}{(x+1)^{2}\left(x^{2}+2 x+2\right)}=\frac{A_{1}}{x+1}+\frac{A_{2}}{(... | -\frac{2}{x+1}+\frac{1}{2}\cdot\ln(x^{2}+2x+2)+\operatorname{arctg}(x+1)+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,902 |
## Problem Statement
Find the indefinite integral:
$$
\int \frac{2 x^{3}+11 x^{2}+16 x+10}{(x+2)^{2}\left(x^{2}+2 x+3\right)} d x
$$ | ## Solution
$$
\int \frac{2 x^{3}+11 x^{2}+16 x+10}{(x+2)^{2}\left(x^{2}+2 x+3\right)} d x=
$$
We decompose the proper rational fraction into partial fractions using the method of undetermined coefficients:
$$
\begin{aligned}
& \frac{2 x^{3}+11 x^{2}+16 x+10}{(x+2)^{2}\left(x^{2}+2 x+3\right)}=\frac{A_{1}}{x+2}+\fra... | -\frac{2}{x+2}+\ln(x^{2}+2x+3)-\frac{1}{\sqrt{2}}\cdot\operatorname{arctg}(\frac{x+1}{\sqrt{2}})+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,903 |
## Problem Statement
Find the indefinite integral:
$$
\int \frac{3 x^{3}+6 x^{2}+5 x-1}{(x+1)^{2}\left(x^{2}+2\right)} d x
$$ | ## Solution
$$
\int \frac{3 x^{3}+6 x^{2}+5 x-1}{(x+1)^{2}\left(x^{2}+2\right)} d x=
$$
We decompose the proper rational fraction into partial fractions using the method of undetermined coefficients:
$$
\begin{aligned}
& \frac{3 x^{3}+6 x^{2}+5 x-1}{(x+1)^{2}\left(x^{2}+2\right)}=\frac{A_{1}}{x+1}+\frac{A_{2}}{(x+1)... | \frac{1}{x+1}+\frac{3}{2}\cdot\ln(x^{2}+2)+\frac{1}{\sqrt{2}}\cdot\operatorname{arctg}\frac{x}{\sqrt{2}}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,904 |
## Problem Statement
Find the indefinite integral:
$$
\int \frac{x^{3}+9 x^{2}+21 x+21}{(x+3)^{2}\left(x^{2}+3\right)} d x
$$ | ## Solution
$$
\int \frac{x^{3}+9 x^{2}+21 x+21}{(x+3)^{2}\left(x^{2}+3\right)} d x=
$$
We decompose the proper rational fraction into partial fractions using the method of undetermined coefficients:
$$
\begin{aligned}
& \frac{x^{3}+9 x^{2}+21 x+21}{(x+3)^{2}\left(x^{2}+3\right)}=\frac{A_{1}}{x+3}+\frac{A_{2}}{(x+3)... | -\frac{1}{x+3}+\frac{1}{2}\cdot\ln(x^{2}+3)+\frac{2}{\sqrt{3}}\cdot\operatorname{arctg}\frac{x}{\sqrt{3}}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,905 |
## Problem Statement
Find the indefinite integral:
$$
\int \frac{x^{3}+6 x^{2}+8 x+8}{(x+2)^{2}\left(x^{2}+4\right)} d x
$$ | ## Solution
$$
\int \frac{x^{3}+6 x^{2}+8 x+8}{(x+2)^{2}\left(x^{2}+4\right)} d x=
$$
Decompose the proper rational fraction into partial fractions using the method of undetermined coefficients:
$$
\begin{aligned}
& \frac{x^{3}+6 x^{2}+8 x+8}{(x+2)^{2}\left(x^{2}+4\right)}=\frac{A_{1}}{x+2}+\frac{A_{2}}{(x+2)^{2}}+\... | -\frac{1}{x+2}+\frac{1}{2}\cdot\ln(x^{2}+4)+\frac{1}{2}\cdot\operatorname{arctg}\frac{x}{2}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,906 |
## Problem Statement
Find the indefinite integral:
$$
\int \frac{x^{3}+5 x^{2}+12 x+4}{(x+2)^{2}\left(x^{2}+4\right)} d x
$$ | ## Solution
$$
\int \frac{x^{3}+5 x^{2}+12 x+4}{(x+2)^{2}\left(x^{2}+4\right)} d x=
$$
We decompose the proper rational fraction into partial fractions using the method of undetermined coefficients:
$$
\begin{aligned}
& \frac{x^{3}+5 x^{2}+12 x+4}{(x+2)^{2}\left(x^{2}+4\right)}=\frac{A_{1}}{x+2}+\frac{A_{2}}{(x+2)^{... | \frac{1}{x+2}+\frac{1}{2}\cdot\ln(x^{2}+4)+\operatorname{arctg}\frac{x}{2}+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,907 |
## Problem Statement
Find the indefinite integral:
$$
\int \frac{2 x^{3}-4 x^{2}-16 x-12}{(x-1)^{2}\left(x^{2}+4 x+5\right)} d x
$$ | ## Solution
$$
\int \frac{2 x^{3}-4 x^{2}-16 x-12}{(x-1)^{2}\left(x^{2}+4 x+5\right)} d x=
$$
We decompose the proper rational fraction into partial fractions using the method of undetermined coefficients:
$$
\begin{aligned}
& \frac{2 x^{3}-4 x^{2}-16 x-12}{(x-1)^{2}\left(x^{2}+4 x+5\right)}=\frac{A_{1}}{x-1}+\frac{... | \frac{3}{x-1}+\ln(x^{2}+4x+5)-\operatorname{arctg}(x+2)+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,908 |
## Problem Statement
Find the indefinite integral:
$$
\int \frac{-3 x^{3}+13 x^{2}-13 x+1}{(x-2)^{2}\left(x^{2}-x+1\right)} d x
$$ | ## Solution
$$
\int \frac{-3 x^{3}+13 x^{2}-13 x+1}{(x-2)^{2}\left(x^{2}-x+1\right)} d x=
$$
Decompose the proper rational fraction into partial fractions using the method of undetermined coefficients:
$$
\begin{aligned}
& \frac{-3 x^{3}+13 x^{2}-13 x+1}{(x-2)^{2}\left(x^{2}-x+1\right)}=\frac{A_{1}}{x-2}+\frac{A_{2}... | -\frac{1}{x-2}-\frac{3}{2}\cdot\ln(x^{2}-x+1)-\sqrt{3}\cdot\operatorname{arctg}(\frac{2x-1}{\sqrt{3}})+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,909 |
## Problem Statement
Find the indefinite integral:
$$
\int \frac{x^{3}+2 x^{2}+10 x}{(x+1)^{2}\left(x^{2}-x+1\right)} d x
$$ | ## Solution
$$
\int \frac{x^{3}+2 x^{2}+10 x}{(x+1)^{2}\left(x^{2}-x+1\right)} d x=
$$
We decompose the proper rational fraction into partial fractions using the method of undetermined coefficients:
$$
\begin{aligned}
& \frac{x^{3}+2 x^{2}+10 x}{(x+1)^{2}\left(x^{2}-x+1\right)}=\frac{A_{1}}{x+1}+\frac{A_{2}}{(x+1)^{... | \frac{3}{x+1}+\frac{1}{2}\cdot\ln(x^{2}-x+1)+\frac{7}{\sqrt{3}}\cdot\operatorname{arctg}(\frac{2x-1}{\sqrt{3}})+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,910 |
## Problem Statement
Find the indefinite integral:
$$
\int \frac{3 x^{3}+x+46}{(x-1)^{2}\left(x^{2}+9\right)} d x
$$ | ## Solution
$$
\int \frac{3 x^{3}+x+46}{(x-1)^{2}\left(x^{2}+9\right)} d x=
$$
We decompose the proper rational fraction into partial fractions using the method of undetermined coefficients:
$$
\begin{aligned}
& \frac{3 x^{3}+x+46}{(x-1)^{2}\left(x^{2}+9\right)}=\frac{A_{1}}{x-1}+\frac{A_{2}}{(x-1)^{2}}+\frac{B x+C}... | -\frac{5}{x-1}+\frac{3}{2}\cdot\ln(x^{2}+9)+\frac{1}{3}\cdot\operatorname{arctg}(\frac{x}{3})+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,911 |
## Problem Statement
Find the indefinite integral:
$$
\int \frac{4 x^{3}+24 x^{2}+20 x-28}{(x+3)^{2}\left(x^{2}+2 x+2\right)} d x
$$ | ## Solution
$$
\int \frac{4 x^{3}+24 x^{2}+20 x-28}{(x+3)^{2}\left(x^{2}+2 x+2\right)} d x=
$$
We decompose the proper rational fraction into partial fractions using the method of undetermined coefficients:
$$
\begin{aligned}
& \frac{4 x^{3}+24 x^{2}+20 x-28}{(x+3)^{2}\left(x^{2}+2 x+2\right)}=\frac{A_{1}}{x+3}+\fra... | -\frac{4}{x+3}+2\cdot\ln(x^{2}+2x+2)-8\cdot\operatorname{arctg}(x+1)+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,912 |
## Problem Statement
Find the indefinite integral:
$$
\int \frac{2 x^{3}+3 x^{2}+3 x+2}{\left(x^{2}+x+1\right)\left(x^{2}+1\right)} d x
$$ | ## Solution
$$
\int \frac{2 x^{3}+3 x^{2}+3 x+2}{\left(x^{2}+x+1\right)\left(x^{2}+1\right)} d x=
$$
Decompose the proper rational fraction into partial fractions using the method of undetermined coefficients:
$$
\begin{aligned}
& \frac{2 x^{3}+3 x^{2}+3 x+2}{\left(x^{2}+x+1\right)\left(x^{2}+1\right)}=\frac{A x+B}{... | \frac{1}{2}\cdot\ln(x^{2}+x+1)+\frac{1}{\sqrt{3}}\cdot\operatorname{arctg}(\frac{2x+1}{\sqrt{3}})+\frac{1}{2}\cdot\ln(x^{2}+1)+\operatorname{arctg}x+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,913 |
## Problem Statement
Find the indefinite integral:
$$
\int \frac{x^{3}+x+1}{\left(x^{2}+x+1\right)\left(x^{2}+1\right)} d x
$$ | ## Solution
$$
\int \frac{x^{3}+x+1}{\left(x^{2}+x+1\right)\left(x^{2}+1\right)} d x=
$$
Decompose the proper rational fraction into partial fractions using the method of undetermined coefficients:
$$
\begin{aligned}
& \frac{x^{3}+x+1}{\left(x^{2}+x+1\right)\left(x^{2}+1\right)}=\frac{A x+B}{x^{2}+x+1}+\frac{C x+D}{... | \ln(x^{2}+x+1)-\frac{1}{2}\cdot\ln(x^{2}+1)+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,914 |
## Problem Statement
Find the indefinite integral:
$$
\int \frac{x^{2}+x+3}{\left(x^{2}+x+1\right)\left(x^{2}+1\right)} d x
$$ | ## Solution
$$
\int \frac{x^{2}+x+3}{\left(x^{2}+x+1\right)\left(x^{2}+1\right)} d x=
$$
Decompose the proper rational fraction into partial fractions using the method of undetermined coefficients:
$$
\begin{aligned}
& \frac{x^{2}+x+3}{\left(x^{2}+x+1\right)\left(x^{2}+1\right)}=\frac{A x+B}{x^{2}+x+1}+\frac{C x+D}{... | \ln(x^{2}+x+1)+\frac{2}{\sqrt{3}}\cdot\operatorname{arctg}(\frac{2x+1}{\sqrt{3}})-\ln(x^{2}+1)+\operatorname{arctg}x+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,915 |
## Problem Statement
Find the indefinite integral:
$$
\int \frac{2 x^{3}+4 x^{2}+2 x+2}{\left(x^{2}+x+1\right)\left(x^{2}+x+2\right)} d x
$$ | ## Solution
$$
\int \frac{2 x^{3}+4 x^{2}+2 x+2}{\left(x^{2}+x+1\right)\left(x^{2}+x+2\right)} d x=
$$
Decompose the proper rational fraction into partial fractions using the method of undetermined coefficients:
$$
\begin{aligned}
& \frac{2 x^{3}+4 x^{2}+2 x+2}{\left(x^{2}+x+1\right)\left(x^{2}+x+2\right)}=\frac{A x... | -\ln(x^{2}+x+1)+\frac{2}{\sqrt{3}}\cdot\operatorname{arctg}(\frac{2x+1}{\sqrt{3}})+2\cdot\ln(x^{2}+x+2)+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,916 |
## Problem Statement
Find the indefinite integral:
$$
\int \frac{2 x^{3}+7 x^{2}+7 x+9}{\left(x^{2}+x+1\right)\left(x^{2}+x+2\right)} d x
$$ | ## Solution
$$
\int \frac{2 x^{3}+7 x^{2}+7 x+9}{\left(x^{2}+x+1\right)\left(x^{2}+x+2\right)} d x=
$$
Decompose the proper rational fraction into partial fractions using the method of undetermined coefficients:
$$
\begin{aligned}
& \frac{2 x^{3}+7 x^{2}+7 x+9}{\left(x^{2}+x+1\right)\left(x^{2}+x+2\right)}=\frac{A x... | \frac{8}{\sqrt{3}}\cdot\operatorname{arctg}(\frac{2x+1}{\sqrt{3}})+\ln(x^{2}+x+2)+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,917 |
## Problem Statement
Find the indefinite integral:
$$
\int \frac{4 x^{2}+3 x+4}{\left(x^{2}+1\right)\left(x^{2}+x+1\right)} d x
$$ | ## Solution
$$
\int \frac{4 x^{2}+3 x+4}{\left(x^{2}+1\right)\left(x^{2}+x+1\right)} d x=
$$
Decompose the proper rational fraction into partial fractions using the method of undetermined coefficients:
$$
\begin{aligned}
& \frac{4 x^{2}+3 x+4}{\left(x^{2}+1\right)\left(x^{2}+x+1\right)}=\frac{A x+B}{x^{2}+1}+\frac{C... | 3\cdot\operatorname{arctg}x+\frac{2}{\sqrt{3}}\cdot\operatorname{arctg}(\frac{2x+1}{\sqrt{3}})+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,918 |
## Problem Statement
Find the indefinite integral:
$$
\int \frac{3 x^{3}+4 x^{2}+6 x}{\left(x^{2}+2\right)\left(x^{2}+2 x+2\right)} d x
$$ | ## Solution
$$
\int \frac{3 x^{3}+4 x^{2}+6 x}{\left(x^{2}+2\right)\left(x^{2}+2 x+2\right)} d x=
$$
Decompose the proper rational fraction into partial fractions using the method of undetermined coefficients:
$$
\begin{aligned}
& \frac{3 x^{3}+4 x^{2}+6 x}{\left(x^{2}+2\right)\left(x^{2}+2 x+2\right)}=\frac{A x+B}{... | \ln(x^{2}+2)+\frac{1}{2}\cdot\ln(x^{2}+2x+2)-\operatorname{arctg}(x+1)+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,919 |
## Problem Statement
Find the indefinite integral:
$$
\int \frac{2 x^{2}-x+1}{\left(x^{2}-x+1\right)\left(x^{2}+1\right)} d x
$$ | ## Solution
$$
\int \frac{2 x^{2}-x+1}{\left(x^{2}-x+1\right)\left(x^{2}+1\right)} d x=
$$
Decompose the proper rational fraction into partial fractions using the method of undetermined coefficients:
$$
\begin{aligned}
& \frac{2 x^{2}-x+1}{\left(x^{2}-x+1\right)\left(x^{2}+1\right)}=\frac{A x+B}{x^{2}-x+1}+\frac{C x... | \frac{1}{2}\cdot\ln(x^{2}-x+1)+\frac{1}{\sqrt{3}}\cdot\operatorname{arctg}(\frac{2x-1}{\sqrt{3}})-\frac{1}{2}\cdot\ln(x^{2}+1)+\operatorname{arctg}x+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,920 |
## Problem Statement
Find the indefinite integral:
$$
\int \frac{x^{3}+x^{2}+1}{\left(x^{2}-x+1\right)\left(x^{2}+1\right)} d x
$$ | ## Solution
$$
\int \frac{x^{3}+x^{2}+1}{\left(x^{2}-x+1\right)\left(x^{2}+1\right)} d x=
$$
Decompose the proper rational fraction into partial fractions using the method of undetermined coefficients:
$$
\begin{aligned}
& \frac{x^{3}+x^{2}+1}{\left(x^{2}-x+1\right)\left(x^{2}+1\right)}=\frac{A x+B}{x^{2}-x+1}+\frac... | \frac{1}{2}\cdot\ln(x^{2}-x+1)+\frac{1}{\sqrt{3}}\cdot\operatorname{arctg}(\frac{2x-1}{\sqrt{3}})+\operatorname{arctg}x+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,921 |
## Problem Statement
Find the indefinite integral:
$$
\int \frac{x^{3}+x+1}{\left(x^{2}-x+1\right)\left(x^{2}+1\right)} d x
$$ | ## Solution
$$
\int \frac{x^{3}+x+1}{\left(x^{2}-x+1\right)\left(x^{2}+1\right)} d x=
$$
Decompose the proper rational fraction into partial fractions using the method of undetermined coefficients:
$$
\begin{aligned}
& \frac{x^{3}+x+1}{\left(x^{2}-x+1\right)\left(x^{2}+1\right)}=\frac{A x+B}{x^{2}-x+1}+\frac{C x+D}{... | \frac{2}{\sqrt{3}}\cdot\operatorname{arctg}(\frac{2x-1}{\sqrt{3}})+\frac{1}{2}\cdot\ln(x^{2}+1)+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,922 |
## Problem Statement
Find the indefinite integral:
$$
\int \frac{2 x^{3}+2 x+1}{\left(x^{2}-x+1\right)\left(x^{2}+1\right)} d x
$$ | ## Solution
$$
\int \frac{2 x^{3}+2 x+1}{\left(x^{2}-x+1\right)\left(x^{2}+1\right)} d x=
$$
Decompose the proper rational fraction into partial fractions using the method of undetermined coefficients:
$$
\begin{aligned}
& \frac{2 x^{3}+2 x+1}{\left(x^{2}-x+1\right)\left(x^{2}+1\right)}=\frac{A x+B}{x^{2}-x+1}+\frac... | \frac{1}{2}\cdot\ln|x^{2}-x+1|+\sqrt{3}\cdot\operatorname{arctg}(\frac{2x-1}{\sqrt{3}})+\ln|x^{2}+1|+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,923 |
## Problem Statement
Find the indefinite integral:
$$
\int \frac{x^{3}+2 x^{2}+x+1}{\left(x^{2}+x+1\right)\left(x^{2}+1\right)} d x
$$ | ## Solution
$$
\int \frac{x^{3}+2 x^{2}+x+1}{\left(x^{2}+x+1\right)\left(x^{2}+1\right)} d x=
$$
Decompose the proper rational fraction into partial fractions using the method of undetermined coefficients:
$$
\begin{aligned}
& \frac{x^{3}+2 x^{2}+x+1}{\left(x^{2}+x+1\right)\left(x^{2}+1\right)}=\frac{A x+B}{x^{2}+x+... | \frac{2}{\sqrt{3}}\cdot\operatorname{arctg}(\frac{2x+1}{\sqrt{3}})+\frac{1}{2}\cdot\ln(x^{2}+1)+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,924 |
## Problem Statement
Find the indefinite integral:
$$
\int \frac{x+4}{\left(x^{2}+x+2\right)\left(x^{2}+2\right)} d x
$$ | ## Solution
$$
\int \frac{x+4}{\left(x^{2}+x+2\right)\left(x^{2}+2\right)} d x=
$$
Decompose the proper rational fraction into partial fractions using the method of undetermined coefficients:
$$
\begin{aligned}
& \frac{x+4}{\left(x^{2}+x+2\right)\left(x^{2}+2\right)}=\frac{A x+B}{x^{2}+x+2}+\frac{C x+D}{x^{2}+2}= \\... | \ln(x^{2}+x+2)-\ln(x^{2}+2)+\frac{1}{\sqrt{2}}\cdot\operatorname{arctg}(\frac{x}{\sqrt{2}})+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,925 |
## Problem Statement
Find the indefinite integral:
$$
\int \frac{2 x^{3}+2 x^{2}+2 x+1}{\left(x^{2}+x+1\right)\left(x^{2}+1\right)} d x
$$ | ## Solution
$$
\int \frac{2 x^{3}+2 x^{2}+2 x+1}{\left(x^{2}+x+1\right)\left(x^{2}+1\right)} d x=
$$
Decompose the proper rational fraction into partial fractions using the method of undetermined coefficients:
$$
\begin{aligned}
& \frac{2 x^{3}+2 x^{2}+2 x+1}{\left(x^{2}+x+1\right)\left(x^{2}+1\right)}=\frac{A x+B}{... | \frac{1}{2}\cdot\ln(x^{2}+x+1)+\frac{1}{\sqrt{3}}\cdot\operatorname{arctg}(\frac{2x+1}{\sqrt{3}})+\frac{1}{2}\cdot\ln(x^{2}+1)+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,926 |
## Problem Statement
Find the indefinite integral:
$$
\int \frac{3 x^{3}+7 x^{2}+12 x+6}{\left(x^{2}+x+3\right)\left(x^{2}+2 x+3\right)} d x
$$ | ## Solution
$$
\int \frac{3 x^{3}+7 x^{2}+12 x+6}{\left(x^{2}+x+3\right)\left(x^{2}+2 x+3\right)} d x=
$$
Decompose the proper rational fraction into partial fractions using the method of undetermined coefficients:
$$
\begin{aligned}
& \frac{3 x^{3}+7 x^{2}+12 x+6}{\left(x^{2}+x+3\right)\left(x^{2}+2 x+3\right)}=\fr... | \ln(x^{2}+x+3)+\frac{1}{2}\cdot\ln(x^{2}+2x+3)+C | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,927 |
## Task Condition
Write the decomposition of vector $x$ in terms of vectors $p, q, r$:
$x=\{8 ; 9 ; 4\}$
$p=\{1 ; 0 ; 1\}$
$q=\{0 ;-2 ; 1\}$
$r=\{1 ; 3 ; 0\}$ | ## Solution
The desired decomposition of vector $x$ is:
$x=\alpha \cdot p+\beta \cdot q+\gamma \cdot r$
Or in the form of a system:
$$
\left\{\begin{array}{l}
\alpha \cdot p_{1}+\beta \cdot q_{1}+\gamma \cdot r_{1}=x_{1} \\
\alpha \cdot p_{2}+\beta \cdot q_{2}+\gamma \cdot r_{2}=x_{2} \\
\alpha \cdot p_{3}+\beta \c... | 7p-3q+r | Algebra | math-word-problem | Yes | Yes | olympiads | false | 45,928 |
## Problem Statement
Are the vectors $c_{1 \text { and }} c_{2}$, constructed from vectors $a \text{ and } b$, collinear?
$a=\{2 ;-1 ; 4\}$
$b=\{3 ;-7 ;-6\}$
$c_{1}=2 a-3 b$
$c_{2}=3 a-2 b$ | ## Solution
Vectors are collinear if there exists a number $\gamma$ such that $c_{1}=\gamma \cdot c_{2}$. That is, vectors are collinear if their coordinates are proportional.
We find:
$$
\begin{aligned}
& c_{1}=2 a-3 b=\{2 \cdot 2-3 \cdot 3 ; 2 \cdot(-1)-3 \cdot(-7) ; 2 \cdot 4-3 \cdot(-6)\}=\{-5 ; 19 ; 26\} \\
& c... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 45,929 |
## Problem Statement
Find the cosine of the angle between vectors $\overrightarrow{A B}$ and $\overrightarrow{A C}$.
$A(2 ; 3 ; 2), B(-1 ;-3 ;-1), C(-3 ;-7 ;-3)$ | ## Solution
Let's find $\overrightarrow{A B}$ and $\overrightarrow{A C}$:
$\overrightarrow{A B}=(-1-2 ;-3-3 ;-1-2)=(-3 ;-6 ;-3)$
$\overrightarrow{A C}=(-3-2 ;-7-3 ;-3-2)=(-5 ;-10 ;-5)$
We find the cosine of the angle $\phi$ between the vectors $\overrightarrow{A B}$ and $\overrightarrow{A C}$:
$\cos (\overrightarr... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 45,930 |
## Task Condition
Calculate the area of the parallelogram constructed on vectors $a$ and $b$.
$a=3 p+4 q$
$b=q-p$
$|p|=2.5$
$|q|=2$
$(\widehat{p, q})=\frac{\pi}{2}$ | ## Solution
The area of the parallelogram constructed on vectors $a$ and $b$ is numerically equal to the modulus of their vector product:
$S=|a \times b|$
We compute $a \times b$ using the properties of the vector product:
$a \times b=(3 p+4 q) \times(q-p)=3 \cdot p \times q+3 \cdot(-1) \cdot p \times p+4 \cdot q \... | 35 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 45,931 |
## Task Condition
Are the vectors $a, b$ and $c$ coplanar?
$a=\{3 ; 4 ; 2\}$
$b=\{1 ; 1 ; 0\}$
$c=\{8 ; 11 ; 6\}$ | ## Solution
For three vectors to be coplanar (lie in the same plane or parallel planes), it is necessary and sufficient that their scalar triple product $(a, b, c)$ be equal to zero.
$$
\begin{aligned}
& (a, b, c)=\left|\begin{array}{ccc}
3 & 4 & 2 \\
1 & 1 & 0 \\
8 & 11 & 6
\end{array}\right|= \\
& =3 \cdot\left|\be... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 45,932 |
## Problem Statement
Calculate the volume of the tetrahedron with vertices at points \( A_{1}, A_{2}, A_{3}, A_{4} \) and its height dropped from vertex \( A_{4} \) to the face \( A_{1} A_{2} A_{3} \).
\( A_{1}(1 ; 0 ; 2) \)
\( A_{2}(1 ; 2 ;-1) \)
\( A_{3}(2 ;-2 ; 1) \)
\( A_{4}(2 ; 1 ; 0) \) | ## Solution
From vertex $A_{1}$, we draw vectors:
$$
\begin{aligned}
& \overrightarrow{A_{1} A_{2}}=\{1-1 ; 2-0 ;-1-2\}=\{0 ; 2 ;-3\} \\
& \overrightarrow{A_{1} A_{3}}=\{2-1 ;-2-0 ; 1-2\}=\{1 ;-2 ;-1\} \\
& \overrightarrow{A_{1} A_{4}}=\{2-1 ; 1-0 ; 0-2\}=\{1 ; 1 ;-2\}
\end{aligned}
$$
According to the geometric mea... | 1\frac{1}{6},\sqrt{\frac{7}{11}} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 45,933 |
## Problem Statement
Find the distance from point $M_{0}$ to the plane passing through three points $M_{1}, M_{2}, M_{3}$.
$M_{1}(5 ; 2 ; 0)$
$M_{2}(2 ; 5 ; 0)$
$M_{3}(1 ; 2 ; 4)$
$M_{0}(-3 ;-6 ;-8)$ | ## Solution
Find the equation of the plane passing through three points $M_{1}, M_{2}, M_{3}$:
$$
\left|\begin{array}{lll}
x-5 & y-2 & z-0 \\
2-5 & 5-2 & 0-0 \\
1-5 & 2-2 & 4-0
\end{array}\right|=0
$$
Perform transformations:
$$
\begin{aligned}
& \left|\begin{array}{ccc}
x-5 & y-2 & z \\
-3 & 3 & 0 \\
-4 & 0 & 4
\e... | 8\sqrt{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 45,934 |
## Problem Statement
Write the equation of the plane passing through point $A$ and perpendicular to vector $\overrightarrow{B C}$.
$A(-4, -2, 5)$
$B(3, -3, -7)$
$C(9, 3, -7)$ | ## Solution
Let's find the vector $\overrightarrow{BC}:$
$\overrightarrow{BC}=\{9-3 ; 3-(-3) ;-7-(-7)\}=\{6 ; 6 ; 0\}$
Since the vector $\overrightarrow{BC}$ is perpendicular to the desired plane, it can be taken as the normal vector. Therefore, the equation of the plane will be:
$6 \cdot(x-(-4))+6 \cdot(y-(-2))+0 ... | x+y+6=0 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 45,935 |
## Task Condition
Find the angle between the planes:
$2 x-z+5=0$
$2 x+3 y-7=0$ | ## Solution
The dihedral angle between planes is equal to the angle between their normal vectors. The normal vectors of the given planes:
$\overrightarrow{n_{1}}=\{2 ; 0 ;-1\}$
$\overrightarrow{n_{2}}=\{2 ; 3 ; 0\}$
The angle $\phi_{\text {between the planes is determined by the formula: }}$
$$
\begin{aligned}
& \... | \arccos\frac{4}{\sqrt{65}}\approx60^{0}15^{\}18^{\\} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 45,936 |
## Problem Statement
Find the coordinates of point $A$, which is equidistant from points $B$ and $C$.
$A(x ; 0 ; 0)$
$B(0 ; 1 ; 3)$
$C(2 ; 0 ; 4)$ | ## Solution
Let's find the distances $A B$ and $A C:$
$$
\begin{aligned}
& A B=\sqrt{(0-x)^{2}+(1-0)^{2}+(3-0)^{2}}=\sqrt{x^{2}+1+9}=\sqrt{x^{2}+10} \\
& A C=\sqrt{(2-x)^{2}+(0-0)^{2}+(4-0)^{2}}=\sqrt{4-4 x+x^{2}+0+16}=\sqrt{x^{2}-4 x+20}
\end{aligned}
$$
Since according to the problem $A B=A C$, then
$$
\sqrt{x^{2... | A(2.5;0;0) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 45,937 |
## Problem Statement
Let $k$ be the coefficient of similarity transformation with the center at the origin. Is it true that point $A$ belongs to the image of plane $a$?
$A(3 ; 5 ; 2)$
$a: 5x - 3y + z - 4 = 0$
$k = \frac{1}{2}$ | ## Solution
When transforming similarity with the center at the origin of the plane
$a: A x+B y+C z+D=0_{\text{and coefficient }} k$ transitions to the plane
$a^{\prime}: A x+B y+C z+k \cdot D=0$. We find the image of the plane $a$:
$a^{\prime}: 5 x-3 y+z-2=0$
Substitute the coordinates of point $A$ into the equat... | 0 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 45,938 |
## Task Condition
Write the canonical equations of the line.
$$
\begin{aligned}
& x+5 y-z+11=0 \\
& x-y+2 z-1=0
\end{aligned}
$$ | ## Solution
Canonical equations of a line:
$\frac{x-x_{0}}{m}=\frac{y-y_{0}}{n}=\frac{z-z_{0}}{p}$
where $\left(x_{0} ; y_{0} ; z_{0}\right)_{\text {- coordinates of some point on the line, and }} \vec{s}=\{m ; n ; p\}$ - its direction vector.
Since the line belongs to both planes simultaneously, its direction vecto... | \frac{x+1}{9}=\frac{y+2}{-3}=\frac{z}{-6} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 45,939 |
## Problem Statement
Find the point of intersection of the line and the plane.
$\frac{x-5}{-1}=\frac{y+3}{5}=\frac{z-1}{2}$
$3 x+7 y-5 z-11=0$ | ## Solution
Let's write the parametric equations of the line.
$$
\begin{aligned}
& \frac{x-5}{-1}=\frac{y+3}{5}=\frac{z-1}{2}=t \Rightarrow \\
& \left\{\begin{array}{l}
x=5-t \\
y=-3+5 t \\
z=1+2 t
\end{array}\right.
\end{aligned}
$$
Substitute into the equation of the plane:
$3(5-t)+7(-3+5 t)-5(1+2 t)-11=0$
$15-3... | (4;2;3) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 45,940 |
## Problem Statement
Find the point $M^{\prime}$ symmetric to the point $M$ with respect to the plane.
$M(1 ; 1 ; 1)$
$x+4 y+3 z+5=0$ | ## Solution
Let's find the equation of the line that is perpendicular to the given plane and passes through point $M$.
Since the line is perpendicular to the given plane, we can take the normal vector of the plane as its direction vector:
$\vec{s}=\vec{n}=\{1 ; 4 ; 3\}$
Then the equation of the desired line is:
$\... | M^{\}(0;-3;-2) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 45,941 |
## Problem Statement
Based on the definition of the derivative, find $f^{\prime}(0)$:
$$
f(x)=\left\{\begin{array}{c}
x^{2} e^{|x|} \sin \frac{1}{x^{2}}, x \neq 0 \\
0, x=0
\end{array}\right.
$$ | ## Solution
By definition, the derivative at the point $x=0$:
$f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}$
Based on the definition, we find:
$$
\begin{aligned}
& f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}=\lim _{\Delta x \rightarrow 0} \fra... | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,942 |
## Condition of the problem
To derive the equation of the tangent line to the given curve at the point with abscissa $x_{0}$.
$$
y=\frac{3 x-2 x^{3}}{3}, x_{0}=1
$$ | ## Solution
Let's find $y^{\prime}:$
$y^{\prime}=\left(\frac{3 x-2 x^{3}}{3}\right)^{\prime}=\frac{3-6 x^{2}}{3}=1-2 x^{2}$
Then:
$y_{0}^{\prime}=y^{\prime}\left(x_{0}\right)=1-2 \cdot 1^{2}=-1$
Since the function $y^{\prime}$ at the point $x_{0}$ has a finite derivative, the equation of the tangent line is:
$y-y... | -x+1\frac{1}{3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,943 |
## Problem Statement
Find the differential $d y$.
$y=\sqrt{3+x^{2}}-x \ln \left|x+\sqrt{3+x^{2}}\right|$ | ## Solution
$$
\begin{aligned}
& d y=y^{\prime} \cdot d x=\left(\sqrt{3+x^{2}}-x \ln \left|x+\sqrt{3+x^{2}}\right|\right)^{\prime} d x= \\
& =\left(\frac{1}{2 \sqrt{3+x^{2}}} \cdot 2 x-\left(\ln \left|x+\sqrt{3+x^{2}}\right|+x \cdot \frac{1}{x+\sqrt{3+x^{2}}} \cdot\left(1+\frac{1}{2 \sqrt{3+x^{2}}} \cdot 2 x\right)\ri... | -\ln|x+\sqrt{3+x^{2}}|\cdot | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,944 |
## Task Condition
Approximately calculate using the differential.
$y=\sqrt[3]{3 x+\cos x}, x=0.01$ | ## Solution
If the increment $\Delta x = x - x_{0}$ of the argument $x$ is small in absolute value, then
$f(x) = f\left(x_{0} + \Delta x\right) \approx f\left(x_{0}\right) + f^{\prime}\left(x_{0}\right) \cdot \Delta x$
Choose:
$x_{0} = 0$
Then:
$\Delta x = 0.01$
Calculate:
$y(0) = \sqrt[3]{3 \cdot 0 + \cos 0} =... | 1.01 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,945 |
## Task Condition
Find the derivative.
$y=\frac{x^{2}+2}{2 \sqrt{1-x^{4}}}$ | ## Solution
$y^{\prime}=\left(\frac{x^{2}+2}{2 \sqrt{1-x^{4}}}\right)^{\prime}=\frac{2 x \cdot \sqrt{1-x^{4}}-\left(x^{2}+2\right) \cdot \frac{1}{2 \sqrt{1-x^{4}}} \cdot\left(-4 x^{3}\right)}{2\left(1-x^{4}\right)}=$
$=\frac{x \cdot\left(1-x^{4}\right)+x^{3}\left(x^{2}+2\right)}{\left(1-x^{4}\right) \sqrt{1-x^{4}}}=\f... | \frac{2x^{3}+x}{(1-x^{4})\sqrt{1-x^{4}}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,946 |
## Task Condition
Find the derivative.
$y=-\frac{e^{3 x}}{3 \operatorname{sh}^{3} x}$ | ## Solution
$y^{\prime}=\left(-\frac{e^{3 x}}{3 \operatorname{sh}^{3} x}\right)^{\prime}=-\frac{3 e^{3 x} \cdot \operatorname{sh}^{3} x-e^{3 x} \cdot 3 \operatorname{sh}^{2} x \cdot \operatorname{ch} x}{3 \operatorname{sh}^{6} x}=$
$=-\frac{e^{3 x} \cdot \operatorname{sh} x-e^{3 x} \cdot \operatorname{ch} x}{\operato... | \frac{e^{3x}\cdot(\operatorname{ch}x-\operatorname{sh}x)}{\operatorname{sh}^{4}x} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,947 |
Condition of the problem
Find the derivative.
$y=\ln \left(\frac{\ln x}{\sin \frac{1}{x}}\right)$ | ## Solution
$y^{\prime}=\left(\ln \left(\frac{\ln x}{\sin \frac{1}{x}}\right)\right)^{\prime}=\frac{\sin \frac{1}{x}}{\ln x} \cdot\left(\frac{\ln x}{\sin \frac{1}{x}}\right)^{\prime}=$
$=\frac{\sin \frac{1}{x}}{\ln x} \cdot \frac{\frac{1}{x} \cdot \sin \frac{1}{x}-\ln x \cdot \cos \frac{1}{x}}{\sin ^{2} \frac{1}{x}}=... | \frac{\sin\frac{1}{x}-x\cdot\lnx\cdot\cos\frac{1}{x}}{x\cdot\lnx\cdot\sin\frac{1}{x}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,948 |
## Task Condition
Find the derivative.
$$
y=\sin \sqrt[3]{\operatorname{tg} 2}-\frac{\cos ^{2} 28 x}{56 \sin 56 x}
$$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(\sin \sqrt[3]{\tan 2}-\frac{\cos ^{2} 28 x}{56 \sin 56 x}\right)^{\prime}=0-\left(\frac{\cos ^{2} 28 x}{56 \sin 56 x}\right)^{\prime}= \\
& =-\left(\frac{\cos ^{2} 28 x}{112 \sin 28 x \cdot \cos 28 x}\right)^{\prime}=-\left(\frac{\cos 28 x}{112 \sin 28 x}\right)^{\prim... | \frac{1}{4\sin^{2}28x} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,949 |
## Problem Statement
Find the derivative.
$$
y=\left(2 x^{2}-x+\frac{1}{2}\right) \operatorname{arctg} \frac{x^{2}-1}{x \sqrt{3}}-\frac{x^{3}}{2 \sqrt{3}}-\frac{\sqrt{3}}{2} \cdot x
$$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(\left(2 x^{2}-x+\frac{1}{2}\right) \operatorname{arctg} \frac{x^{2}-1}{x \sqrt{3}}-\frac{x^{3}}{2 \sqrt{3}}-\frac{\sqrt{3}}{2} \cdot x\right)^{\prime}= \\
& =(4 x-1) \operatorname{arctg} \frac{x^{2}-1}{x \sqrt{3}}+\left(2 x^{2}-x+\frac{1}{2}\right) \cdot \frac{1}{1+\le... | (4x-1)\operatorname{arctg}\frac{x^{2}-1}{x\sqrt{3}}+\frac{\sqrt{3}(x^{2}+1)(3x^{2}-2x-x^{4})}{2(x^{4}+x^{2}+1)} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,950 |
## Problem Statement
Find the derivative.
$$
y=\frac{\operatorname{sh} x}{2 \operatorname{ch}^{2} x}+\frac{1}{2} \cdot \operatorname{arctg}(\operatorname{sh} x)
$$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(\frac{\operatorname{sh} x}{2 \operatorname{ch}^{2} x}+\frac{1}{2} \cdot \operatorname{arctg}(\operatorname{sh} x)\right)^{\prime}= \\
& =\frac{\operatorname{ch} x \cdot \operatorname{ch}^{2} x-\operatorname{sh} x \cdot 2 \operatorname{ch} x \cdot \operatorname{sh} x}{2... | \frac{1}{\operatorname{ch}^{3}x} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,951 |
## Task Condition
Find the derivative.
$y=\left(x^{8}+1\right)^{\text{th } x}$ | ## Solution
$y=\left(x^{8}+1\right)^{\text{th } x}$
$\ln y=\ln \left(x^{8}+1\right)^{\text{th } x}=\operatorname{th} x \cdot \ln \left(x^{8}+1\right)$
$\frac{y^{\prime}}{y}=\frac{1}{\operatorname{ch}^{2} x} \cdot \ln \left(x^{8}+1\right)+\operatorname{th} x \cdot \frac{1}{x^{8}+1} \cdot 8 x^{7}=\frac{\ln \left(x^{8}... | (x^{8}+1)^{\operatorname{}x}\cdot(\frac{\ln(x^{8}+1)}{\operatorname{ch}^{2}x}+\frac{8x^{7}\cdot\operatorname{}x}{x^{8}+1}) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,952 |
## Problem Statement
Find the derivative.
$y=\ln \left(e^{3 x}+\sqrt{e^{6 x}-1}\right)+\arcsin \left(e^{-3 x}\right)$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(\ln \left(e^{3 x}+\sqrt{e^{6 x}-1}\right)+\arcsin \left(e^{-3 x}\right)\right)^{\prime}= \\
& =\frac{1}{\left(e^{3 x}+\sqrt{e^{6 x}-1}\right)} \cdot\left(3 e^{3 x}+\frac{1}{2 \sqrt{e^{6 x}-1}} \cdot 6 e^{6 x}\right)+\frac{1}{\sqrt{1-e^{-6 x}}} \cdot\left(-3 e^{-3 x}\ri... | 3\sqrt{\frac{e^{3x}-1}{e^{3x}+1}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,953 |
## Task Condition
Find the derivative.
$$
y=\frac{x}{4}\left(10-x^{2}\right) \sqrt{4-x^{2}}+6 \arcsin \frac{x}{2}
$$ | ## Solution
$$
\begin{aligned}
& y^{\prime}=\left(\frac{x}{4}\left(10-x^{2}\right) \sqrt{4-x^{2}}+6 \arcsin \frac{x}{2}\right)^{\prime}= \\
& =\frac{10-3 x^{2}}{4} \cdot \sqrt{4-x^{2}}+\frac{x}{4}\left(10-x^{2}\right) \cdot \frac{1}{2 \sqrt{4-x^{2}}} \cdot(-2 x)+6 \cdot \frac{1}{\sqrt{1-\left(\frac{x}{2}\right)^{2}}} ... | \sqrt{(4-x^{2})^{3}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,954 |
## Problem Statement
Find the derivative.
$$
y=\frac{\cos x}{3(2+\sin x)}+\frac{4}{3 \sqrt{3}} \operatorname{arctg} \frac{2 \operatorname{tg}\left(\frac{x}{2}\right)+1}{\sqrt{3}}
$$ | ## Solution
$y^{\prime}=\left(\frac{\cos x}{3(2+\sin x)}+\frac{4}{3 \sqrt{3}} \operatorname{arctg} \frac{2 \operatorname{tg}\left(\frac{x}{2}\right)+1}{\sqrt{3}}\right)^{\prime}=$
$$
\begin{aligned}
& =\frac{-\sin x \cdot(2+\sin x)-\cos x \cdot \cos x}{3(2+\sin x)^{2}}+\frac{4}{3 \sqrt{3}} \cdot \frac{1}{1+\left(\fra... | \frac{2\sinx+7}{3(2+\sinx)^{2}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,955 |
## Problem Statement
Find the derivative $y_{x}^{\prime}$
$$
\left\{\begin{array}{l}
x=\frac{t^{2} \ln t}{1-t^{2}}+\ln \sqrt{1-t^{2}} \\
y=\frac{t}{\sqrt{1-t^{2}}} \arcsin t+\ln \sqrt{1-t^{2}}
\end{array}\right.
$$ | ## Solution
$x_{t}^{\prime}=\left(\frac{t^{2} \ln t}{1-t^{2}}+\ln \sqrt{1-t^{2}}\right)^{\prime}=$
$=\frac{\left(2 t \ln t+t^{2} \cdot \frac{1}{t}\right) \cdot\left(1-t^{2}\right)-t^{2} \ln t \cdot(-2 t)}{\left(1-t^{2}\right)^{2}}+\frac{1}{\sqrt{1-t^{2}}} \cdot \frac{1}{2 \sqrt{1-t^{2}}} \cdot(-2 t)=$
$=\frac{t \cdo... | \frac{\arcsin\cdot\sqrt{1-^{2}}}{2\cdot\ln} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,956 |
## Problem Statement
Derive the equations of the tangent and normal lines to the curve at the point corresponding to the parameter value $t=t_{0}$.
$$
\begin{aligned}
& \left\{\begin{array}{l}
x=t^{3}+1 \\
y=t^{2}
\end{array}\right. \\
& t_{0}=-2
\end{aligned}
$$ | ## Solution
Since $t_{0}=-2$, then
$x_{0}=(-2)^{3}+1=9$
$y_{0}=(-2)^{2}=4$
Find the derivatives:
$x_{t}^{\prime}=\left(t^{3}+1\right)^{\prime}=3 t^{2}$
$y_{t}^{\prime}=\left(t^{2}\right)^{\prime}=2 t$
$y_{x}^{\prime}=\frac{y_{t}^{\prime}}{x_{t}^{\prime}}=\frac{2 t}{3 t^{2}}=\frac{2}{3 t}$
Then:
$$
y_{0}^{\prim... | -\frac{x}{3}+73x-23 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,957 |
## Task Condition
Find the $n$-th order derivative.
$y=\log _{3}(x+5)$ | ## Solution
$y=\log _{3}(x+5)$
$y^{\prime}=\left(\log _{3}(x+5)\right)^{\prime}=\frac{1}{(x+5) \ln 3}=\frac{1}{\ln 3} \cdot(x+5)^{-1}$
$y^{\prime \prime}=\left(y^{\prime}\right)^{\prime}=\left(\frac{1}{\ln 3} \cdot(x+5)^{-1}\right)^{\prime}=-\frac{1}{\ln 3} \cdot(x+5)^{-2}$
$y^{\prime \prime \prime}=\left(y^{\prime... | y^{(n)}=\frac{(-1)^{n-1}\cdot(n-1)!}{\ln3\cdot(x+5)^{n}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,958 |
## Task Condition
Find the derivative of the specified order.
$y=e^{-x} \cdot(\cos 2 x-3 \sin 2 x), y^{IV}=?$ | ## Solution
$y^{\prime}=\left(e^{-x} \cdot(\cos 2 x-3 \sin 2 x)\right)^{\prime}=$
$=-e^{-x} \cdot(\cos 2 x-3 \sin 2 x)+e^{-x} \cdot(-2 \sin 2 x-6 \cos 2 x)=$
$=-e^{-x} \cdot(\cos 2 x-3 \sin 2 x+2 \sin 2 x+6 \cos 2 x)=$
$=-e^{-x} \cdot(7 \cos 2 x-\sin 2 x)$
$y^{\prime \prime}=\left(y^{\prime}\right)^{\prime}=\left(... | -e^{-x}\cdot(79\cos2x+3\sin2x) | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,959 |
## Problem Statement
Find the second-order derivative $y_{x x}^{\prime \prime}$ of the function given parametrically.
$$
\left\{\begin{array}{l}
x=\sin t - t \cdot \cos t \\
y=\cos t + t \cdot \sin t
\end{array}\right.
$$ | ## Solution
$x_{t}^{\prime}=(\sin t-t \cdot \cos t)^{\prime}=\cos t-\cos t+t \cdot \sin t=t \cdot \sin t$
$y_{t}^{\prime}=(\cos t+t \cdot \sin t)^{\prime}=-\sin t+\sin t+t \cdot \cos t=t \cdot \cos t$
We obtain:
$$
\begin{aligned}
& y_{x}^{\prime}=\frac{y_{t}^{\prime}}{x_{t}^{\prime}}=\frac{t \cdot \cos t}{t \cdot ... | -\frac{1}{\cdot\sin^{3}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,960 |
## Condition of the problem
Prove that $\lim _{n \rightarrow \infty} a_{n}=a_{\text {(specify }} N(\varepsilon)$ ).
$a_{n}=\frac{5 n+1}{10 n-3}, a=\frac{1}{2}$ | ## Solution
By the definition of the limit:
$\forall \varepsilon>0: \exists N(\varepsilon) \in \mathbb{N}: \forall n: n \geq N(\varepsilon):\left|a_{n}-a\right| \\
& \left.\frac{5}{2(10 n-3)} \right\rvert\,
\end{aligned}\right.
$
$\frac{5}{2(10 n-3)}$
$10 n+3>\frac{5}{2 \varepsilon} ;=>$
$n>\frac{1}{10}\left(\frac... | N(\varepsilon)=[\frac{5+14\varepsilon}{20\varepsilon}] | Calculus | proof | Yes | Yes | olympiads | false | 45,962 |
## Problem Statement
Calculate the limit of the numerical sequence:
$\lim _{n \rightarrow \infty} \frac{(n+1)^{4}-(n-1)^{4}}{(n+1)^{3}+(n-1)^{3}}$ | ## Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{(n+1)^{4}-(n-1)^{4}}{(n+1)^{3}+(n-1)^{3}}=\lim _{n \rightarrow \infty} \frac{\left((n+1)^{2}-(n-1)^{2}\right) \cdot\left((n+1)^{2}+(n-1)^{2}\right)}{(n+1)^{3}+(n-1)^{3}}= \\
& =\lim _{n \rightarrow \infty} \frac{\left(n^{2}+2 n+1-n^{2}+2 n-1\right) \c... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 45,963 |
## Problem Statement
Calculate the limit of the numerical sequence:
$\lim _{n \rightarrow \infty} \frac{\sqrt[3]{n^{2}+2}-5 n^{2}}{n-\sqrt{n^{4}-n+1}}$ | ## Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{\sqrt[3]{n^{2}+2}-5 n^{2}}{n-\sqrt{n^{4}-n+1}}=\lim _{n \rightarrow \infty} \frac{\frac{1}{n^{2}}\left(\sqrt[3]{n^{2}+2}-5 n^{2}\right)}{\frac{1}{n^{2}}\left(n-\sqrt{n^{4}-n+1}\right)}= \\
& =\lim _{n \rightarrow \infty} \frac{\sqrt[3]{\frac{1}{n^{4}}... | 5 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,964 |
## Problem Statement
Calculate the limit of the numerical sequence:
$\lim _{n \rightarrow \infty}(n-\sqrt{n(n-1)})$ | ## Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty}(n-\sqrt{n(n-1)})=\lim _{n \rightarrow \infty} \frac{(n-\sqrt{n(n-1)})(n+\sqrt{n(n-1)})}{n+\sqrt{n(n-1)}}= \\
& =\lim _{n \rightarrow \infty} \frac{n^{2}-n(n-1)}{n+\sqrt{n(n-1)}}=\lim _{n \rightarrow \infty} \frac{n^{2}-n^{2}+n}{n+\sqrt{n(n-1)}}= \\
& =\lim ... | \frac{1}{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,965 |
## Problem Statement
Calculate the limit of the numerical sequence:
$\lim _{n \rightarrow \infty} \frac{2+4+6+\ldots+2 n}{1+3+5+\ldots+(2 n-1)}$ | ## Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty} \frac{2+4+6+\ldots+2 n}{1+3+5+\ldots+(2 n-1)}=\lim _{n \rightarrow \infty} \frac{\left(\frac{(2+2 n) n}{2}\right)}{\left(\frac{(1+(2 n-1)) n}{2}\right)}= \\
& =\lim _{n \rightarrow \infty} \frac{(2+2 n) n}{(1+(2 n-1)) n}=\lim _{n \rightarrow \infty} \frac{2... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 45,966 |
## Problem Statement
Calculate the limit of the numerical sequence:
$\lim _{n \rightarrow \infty}\left(\frac{n+4}{n+2}\right)^{n}$ | ## Solution
$$
\begin{aligned}
& \lim _{n \rightarrow \infty}\left(\frac{n+4}{n+2}\right)^{n}=\lim _{n \rightarrow \infty}\left(\frac{n+2+2}{n+2}\right)^{n}= \\
& =\lim _{n \rightarrow \infty}\left(1+\frac{2}{n+2}\right)^{n}=\lim _{n \rightarrow \infty}\left(1+\frac{1}{\left(\frac{n+2}{2}\right)}\right)^{n}=
\end{alig... | e^2 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,967 |
## Condition of the problem
Prove that (find $\delta(\varepsilon)$ ):
$$
\lim _{x \rightarrow-5} \frac{x^{2}+2 x-15}{x+5}=-8
$$ | ## Solution
According to the definition of the limit of a function by Cauchy:
If a function $f: M \subset \mathbb{R} \rightarrow \mathbb{R}$ and $a \in M^{\prime}$ is a limit point of the set $M$. A number $A \in \mathbb{R}$ is called the limit of the function $f$ as $x$ approaches $a (x \rightarrow a)$, if
$\forall... | proof | Calculus | proof | Yes | Yes | olympiads | false | 45,968 |
## Condition of the problem
Prove that the function $f(x)_{\text {is continuous at the point }} x_{0}$ (find $\delta(\varepsilon)$ ):
$f(x)=-5 x^{2}-7, x_{0}=1$ | ## Solution
By definition, the function $f(x)_{\text {is continuous at the point }} x=x_{0}$, if $\forall \varepsilon>0: \exists \delta(\varepsilon)>0:$.
$\left|x-x_{0}\right|0_{\text {there exists such }} \delta(\varepsilon)>0$, that $\left|f(x)-f\left(x_{0}\right)\right|<\varepsilon$ when $\left|x-x_{0}\right|<\del... | proof | Calculus | proof | Yes | Yes | olympiads | false | 45,969 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow-1} \frac{x^{2}+3 x+2}{x^{3}+2 x^{2}-x-2}$ | ## Solution
$$
\begin{aligned}
& \lim _{x \rightarrow-1} \frac{x^{2}+3 x+2}{x^{3}+2 x^{2}-x-2}=\left\{\frac{0}{0}\right\}=\lim _{x \rightarrow-1} \frac{(x+1)(x+2)}{(x+1)\left(x^{2}+x-2\right)}= \\
& =\lim _{x \rightarrow-1} \frac{x+2}{x^{2}+x-2}=\frac{-1+2}{(-1)^{2}-1-2}=\frac{1}{-2}=-\frac{1}{2}
\end{aligned}
$$
## ... | -\frac{1}{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,970 |
## Problem Statement
Calculate the limit of the function:
$$
\lim _{x \rightarrow 0} \frac{\sqrt[3]{8+3 x-x^{2}}-2}{\sqrt[3]{x^{2}+x^{3}}}
$$ | ## Solution
$$
\begin{aligned}
& \lim _{x \rightarrow 0} \frac{\sqrt[3]{8+3 x-x^{2}}-2}{\sqrt[3]{x^{2}+x^{3}}}= \\
& =\lim _{x \rightarrow 0} \frac{\left(\sqrt[3]{8+3 x-x^{2}}-2\right)\left(\sqrt[3]{\left(8+3 x-x^{2}\right)^{2}}+2 \sqrt[3]{8+3 x-x^{2}}+4\right)}{\left(\sqrt[3]{x^{2}+x^{3}}\right)\left(\sqrt[3]{\left(8... | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,971 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0} \frac{1-\cos x}{\left(e^{3 x}-1\right)^{2}}$ | ## Solution
Let's use the substitution of equivalent infinitesimals:
$1-\cos x \sim \frac{x^{2}}{2}$, as $x \rightarrow 0$
$e^{3 x}-1 \sim 3 x$, as $x \rightarrow 0(3 x \rightarrow 0)$
We get:
$\lim _{x \rightarrow 0} \frac{1-\cos x}{\left(e^{3 x}-1\right)^{2}}=\left\{\frac{0}{0}\right\}=\lim _{x \rightarrow 0} \fr... | \frac{1}{18} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,972 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0} \frac{1+\cos (x-\pi)}{\left(e^{3 x}-1\right)^{2}}$ | Solution
$\lim _{x \rightarrow 0} \frac{1+\cos (x-\pi)}{\left(e^{3 x}-1\right)^{2}}=\lim _{x \rightarrow 0} \frac{1+\cos (\pi-x)}{\left(e^{3 x}-1\right)^{2}}=$
Using the reduction formula:
$=\lim _{x \rightarrow 0} \frac{1-\cos x}{\left(e^{3 x}-1\right)^{2}}=$
Using the substitution of equivalent infinitesimals:
$... | \frac{1}{18} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,973 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow \pi} \frac{1-\sin \left(\frac{x}{2}\right)}{\pi-x}$ | ## Solution
Substitution:
$$
\begin{aligned}
& x=y+\pi \Rightarrow y=x-\pi \\
& x \rightarrow \pi \Rightarrow y \rightarrow 0
\end{aligned}
$$
We get:
$$
\begin{aligned}
& \lim _{x \rightarrow \pi} \frac{1-\sin \left(\frac{x}{2}\right)}{\pi-x}=\lim _{y \rightarrow 0} \frac{1-\sin \left(\frac{y+\pi}{2}\right)}{\pi-(... | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,974 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow-1} \frac{\tan(x+1)}{e^{\sqrt[3]{x^{3}-4 x^{2}+6}}-e}$ | ## Solution
Substitution:
$x=y-1 \Rightarrow y=x+1$
$x \rightarrow-1 \Rightarrow y \rightarrow 0$
We get:
$$
\begin{aligned}
& \lim _{x \rightarrow-1} \frac{\tan(x+1)}{e^{\sqrt[3]{x^{3}-4 x^{2}+6}}-e}=\lim _{y \rightarrow 0} \frac{\tan((y-1)+1)}{e^{\sqrt[3]{(y-1)^{3}-4(y-1)^{2}+6}}-e}= \\
& =\lim _{y \rightarrow 0... | \frac{3}{11e} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,975 |
## Problem Statement
Calculate the limit of the function:
$$
\lim _{x \rightarrow 0} \frac{e^{x}-e^{3 x}}{\sin 3 x-\tan 2 x}
$$ | ## Solution
$$
\begin{aligned}
& \lim _{x \rightarrow 0} \frac{e^{x}-e^{3 x}}{\sin 3 x-\tan 2 x}=\lim _{x \rightarrow 0} \frac{\left(e^{x}-1\right)-\left(e^{3 x}-1\right)}{\sin 3 x-\tan 2 x}= \\
& =\lim _{x \rightarrow 0} \frac{\frac{1}{x}\left(\left(e^{x}-1\right)-\left(e^{3 x}-1\right)\right)}{\frac{1}{x}(\sin 3 x-\... | -2 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,976 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow \frac{\pi}{6}} \frac{2 \sin ^{2} x+\sin x-1}{2 \sin ^{2} x-3 \sin x+1}$ | ## Solution
$$
\begin{aligned}
& \lim _{x \rightarrow \frac{\pi}{6}} \frac{2 \sin ^{2} x+\sin x-1}{2 \sin ^{2} x-3 \sin x+1}= \\
& =\lim _{x \rightarrow \frac{\pi}{6}} \frac{2\left(\sin ^{2} x+\frac{1}{2} \cdot \sin x+\frac{1}{16}\right)-\frac{9}{8}}{2\left(\sin ^{2} x-\frac{3}{2} \sin x+\frac{9}{16}\right)-\frac{1}{8... | -3 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,977 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow 0}\left(2-e^{x^{2}}\right)^{\frac{1}{1-\cos \pi x}}$ | ## Solution
$\lim _{x \rightarrow 0}\left(2-e^{x^{2}}\right)^{\frac{1}{1-\cos \pi x}}=$
$=\lim _{x \rightarrow 0}\left(e^{\ln \left(2-e^{x^{2}}\right)}\right)^{\frac{1}{1-\cos \pi x}}=$
$=\lim _{x \rightarrow 0} e^{\ln \left(2-e^{x^{2}}\right) /(1-\cos \pi x)}=$
$=\exp \left\{\lim _{x \rightarrow 0} \frac{\ln \left... | e^{-\frac{2}{\pi^{2}}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,978 |
## Problem Statement
Calculate the limit of the function:
$$
\lim _{x \rightarrow 0}\left(\frac{\operatorname{arctg} 3 x}{x}\right)^{x+2}
$$ | ## Solution
$$
\begin{aligned}
& \lim _{x \rightarrow 0}\left(\frac{\operatorname{arctg} 3 x}{x}\right)^{x+2}=\left(\lim _{x \rightarrow 0} \frac{\operatorname{arctg} 3 x}{x}\right)^{\lim _{x \rightarrow 0} x+2}= \\
& =\left(\lim _{x \rightarrow 0} \frac{\operatorname{arctg} 3 x}{x}\right)^{0+2}=\left(\lim _{x \righta... | 9 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,979 |
## Problem Statement
Calculate the limit of the function:
$\lim _{x \rightarrow \frac{\pi}{2}}\left(\operatorname{ctg}\left(\frac{x}{2}\right)\right)^{\frac{1}{\cos x}}$ | ## Solution
Substitution:
$x=2 y+\frac{\pi}{2} \Rightarrow y=\frac{1}{2}\left(x-\frac{\pi}{2}\right)$
$x \rightarrow \frac{\pi}{2} \Rightarrow y \rightarrow 0$
We get:
$$
\begin{aligned}
& \lim _{x \rightarrow \frac{\pi}{2}}\left(\operatorname{ctg}\left(\frac{x}{2}\right)\right)^{\frac{1}{\cos x}}=\lim _{y \righta... | e | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,980 |
## Problem Statement
Calculate the limit of the function:
$$
\lim _{x \rightarrow 1}\left(\frac{e^{\sin \pi x}-1}{x-1}\right)^{x^{2}+1}
$$ | ## Solution
$$
\begin{aligned}
& \lim _{x \rightarrow 1}\left(\frac{e^{\sin \pi x}-1}{x-1}\right)^{x^{2}+1}=\left(\lim _{x \rightarrow 1} \frac{e^{\sin \pi x}-1}{x-1}\right)^{\lim _{x \rightarrow 1} x^{2}+1}= \\
& =\left(\lim _{x \rightarrow 1} \frac{e^{\sin \pi x}-1}{x-1}\right)^{1^{2}+1}=\left(\lim _{x \rightarrow 1... | \pi^2 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,981 |
Condition of the problem
Calculate the limit of the function:
$\lim _{x \rightarrow 0} \sqrt{\left(e^{\sin x}-1\right) \cos \left(\frac{1}{x}\right)+4 \cos x}$ | ## Solution
Since $\cos \left(\frac{1}{x}\right)_{\text {- is bounded, and }}$
$\lim _{x \rightarrow 0} e^{\sin x}-1=e^{\sin 0}-1=e^{0}-1=1-1=0$, then
$$
\left(e^{\sin x}-1\right) \cos \left(\frac{1}{x}\right) \rightarrow 0 \quad, \text { as } x \rightarrow 0
$$
Therefore:
$\lim _{x \rightarrow 0} \sqrt{\left(e^{\... | 2 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,982 |
## Problem Statement
Calculate the definite integral:
$$
\int_{1}^{64} \frac{1-\sqrt[6]{x}+2 \sqrt[3]{x}}{x+2 \sqrt{x^{3}}+\sqrt[3]{x^{4}}} d x
$$ | ## Solution
$$
\int_{1}^{64} \frac{1-\sqrt[6]{x}+2 \sqrt[3]{x}}{x+2 \sqrt{x^{3}}+\sqrt[3]{x^{4}}} d x=
$$
Substitution:
$$
\begin{aligned}
& x=t^{6}, d x=6 t^{5} d t \\
& x=1 \Rightarrow t=\sqrt[6]{1}=1 \\
& x=64 \Rightarrow t=\sqrt[6]{64}=2
\end{aligned}
$$
We get:
$$
=\int_{1}^{2} \frac{\left(1-t+2 t^{2}\right) ... | 6\ln\frac{4}{3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,984 |
## Problem Statement
Calculate the definite integral:
$$
\int_{6}^{9} \sqrt{\frac{9-2 x}{2 x-21}} d x
$$ | ## Solution
$$
\int_{6}^{9} \sqrt{\frac{9-2 x}{2 x-21}} d x=
$$
Substitution:
$$
\begin{aligned}
& t=\sqrt{\frac{9-2 x}{2 x-21}} \Rightarrow t^{2}=\frac{9-2 x}{2 x-21}=-\frac{2 x-21+12}{2 x-21}=-1-\frac{12}{2 x-21} \Rightarrow \\
& \Rightarrow 21-2 x=\frac{12}{t^{2}+1} \Rightarrow x=-\frac{6}{t^{2}+1}+\frac{21}{2} \... | \pi | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,986 |
## Problem Statement
Calculate the definite integral:
$$
\int_{0}^{5} e^{\sqrt{(5-x) /(5+x)}} \cdot \frac{d x}{(5+x) \sqrt{25-x^{2}}}
$$ | ## Solution
$$
\int_{0}^{5} e^{\sqrt{(5-x) /(5+x)}} \cdot \frac{d x}{(5+x) \sqrt{25-x^{2}}}=
$$
Substitution:
$$
\begin{aligned}
& t=\sqrt{\frac{5-x}{5+x}} \\
& d t=\frac{1}{2} \cdot \sqrt{\frac{5+x}{5-x}} \cdot\left(\frac{5-x}{5+x}\right)^{\prime} \cdot d x=\frac{1}{2} \cdot \sqrt{\frac{5+x}{5-x}} \cdot \frac{-10}{... | \frac{e-1}{5} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,987 |
## Problem Statement
Calculate the definite integral:
$$
\int_{8}^{12} \sqrt{\frac{6-x}{x-14}} d x
$$ | ## Solution
$$
\int_{8}^{12} \sqrt{\frac{6-x}{x-14}} d x=
$$
Substitution:
$$
\begin{aligned}
& t=\sqrt{\frac{6-x}{x-14}} \Rightarrow t^{2}=\frac{6-x}{x-14}=-\frac{x-14+8}{x-14}=-1-\frac{8}{x-14} \Rightarrow \\
& \Rightarrow 14-x=\frac{8}{t^{2}+1} \Rightarrow x=-\frac{8}{t^{2}+1}+14 \\
& d x=\frac{8}{\left(t^{2}+1\r... | \frac{4\pi}{3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,988 |
## Problem Statement
Calculate the definite integral:
$$
\int_{0}^{1} e^{\sqrt{(1-x) /(1+x)}} \cdot \frac{d x}{(1+x) \sqrt{1-x^{2}}}
$$ | ## Solution
$$
\int_{0}^{1} e^{\sqrt{(1-x) /(1+x)}} \cdot \frac{d x}{(1+x) \sqrt{1-x^{2}}}=
$$
Substitution:
$$
\begin{aligned}
& t=\sqrt{\frac{1-x}{1+x}} \\
& d t=\frac{1}{2} \cdot \sqrt{\frac{1+x}{1-x}} \cdot\left(\frac{1-x}{1+x}\right)^{\prime} \cdot d x=\frac{1}{2} \cdot \sqrt{\frac{1+x}{1-x}} \cdot \frac{-2}{(1... | e-1 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,989 |
## Problem Statement
Calculate the definite integral:
$$
\int_{5 / 2}^{10 / 3} \frac{\sqrt{x+2}+\sqrt{x-2}}{(\sqrt{x+2}-\sqrt{x-2})(x-2)^{2}} d x
$$ | ## Solution
$$
\int_{5 / 2}^{10 / 3} \frac{\sqrt{x+2}+\sqrt{x-2}}{(\sqrt{x+2}-\sqrt{x-2})(x-2)^{2}} d x=
$$
Divide the numerator and the denominator by $\sqrt{x-2}$:
$$
=\int_{5 / 2}^{10 / 3} \frac{\sqrt{\frac{x+2}{x-2}}+1}{\left(\sqrt{\frac{x+2}{x-2}}-1\right)(x-2)^{2}} d x=
$$
Substitution:
$$
\begin{aligned}
& ... | \frac{9}{4}+\ln2 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,990 |
## Problem Statement
Calculate the definite integral:
$$
\int_{1}^{8} \frac{5 \sqrt{x+24}}{(x+24)^{2} \cdot \sqrt{x}} d x
$$ | ## Solution
$=\int_{2}^{5} \frac{240 t^{2} dt}{576 \cdot t^{4}}=\frac{15}{36} \int_{2}^{5} t^{-2} dt=-\left.\frac{15}{36 t}\right|_{2} ^{5}=-\frac{15}{36}\left(\frac{1}{5}-\frac{1}{2}\right... | \frac{1}{8} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,991 |
## Problem Statement
Calculate the definite integral:
$$
\int_{1}^{2} \frac{x+\sqrt{3 x-2}-10}{\sqrt{3 x-2}+7} d x
$$ | ## Solution
11.10. $\int_{1}^{2} \frac{x+\sqrt{3 x-2}-10}{\sqrt{3 x-2}+7} d x=$
saucera: $\sqrt{3 x-2}=t \Rightarrow x=\frac{1}{3}\left(t^{2}+2\right)$, $d x=\frac{2}{3} t d t$
$=\frac{2}{3} \int_{1}^{2} \frac{\frac{1}{3}\left(t^{2}+2\right)+t-10}{t+7} t d t=\frac{2}{9} \int_{1}^{2} \frac{t^{3}+3 t^{2}-28 t}{t+7} d ... | -\frac{22}{27} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,992 |
## Problem Statement
Calculate the definite integral:
$$
\int_{6}^{10} \sqrt{\frac{4-x}{x-12}} d x
$$ | ## Solution
$$
\int_{6}^{10} \sqrt{\frac{4-x}{x-12}} d x=
$$
Substitution:
$$
\begin{aligned}
& t=\sqrt{\frac{4-x}{x-12}} \Rightarrow t^{2}=\frac{4-x}{x-12}=-\frac{x-12+8}{x-12}=-1-\frac{8}{x-12} \Rightarrow \\
& \Rightarrow 12-x=\frac{8}{t^{2}+1} \Rightarrow x=-\frac{8}{t^{2}+1}+12 \\
& d x=\frac{8}{\left(t^{2}+1\r... | \frac{4\pi}{3} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,993 |
## Problem Statement
Calculate the definite integral:
$$
\int_{0}^{2} \frac{(4 \sqrt{2-x}-\sqrt{2 x+2}) d x}{(\sqrt{2 x+2}+4 \sqrt{2-x})(2 x+2)^{2}}
$$ | ## Solution
Let's introduce the substitution:
$$
t=\sqrt{\frac{2-x}{2 x+2}}
$$
Then
$$
x=\frac{2-2 t^{2}}{2 t^{2}+1} \Rightarrow 2 x+2=\frac{6}{2 t^{2}+1} \Rightarrow d x=-\frac{12 t}{\left(2 t^{2}+1\right)} d t
$$
When $x=0, t=1$
When $x=2, t=0$
We get:
$$
\begin{aligned}
& \int_{0}^{2} \frac{(4 \sqrt{2-x}-\sq... | \frac{1}{24}\ln5 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,994 |
## Problem Statement
Calculate the definite integral:
$$
\int_{-1 / 2}^{0} \frac{x \cdot d x}{2+\sqrt{2 x+1}}
$$ | ## Solution
$$
\int_{-1 / 2}^{0} \frac{x d x}{2+\sqrt{2 x+1}}=
$$
Perform the substitution:
$$
t^{2}=2 x+1 ; d x=t d t
$$
Recalculate the limits of integration:
$$
\begin{aligned}
& x=-\frac{1}{2} \Rightarrow t=0 \\
& x=0 \Rightarrow t=1
\end{aligned}
$$
Then we get:
$$
\begin{aligned}
& \frac{1}{2} \int_{0}^{1}... | \frac{7}{6}-3\ln\frac{3}{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 45,995 |
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