Question

$\int e^{{\tan }^{-1}x}\left(1+x+x^2\right)\cdot d\left({\cot }^{-1}x\right)$ is equal to

• A. $-e^{{\tan }^{-1}x}+C$
• B. $e^{{\tan }^{-1}x}+C$
• C. $-x\cdot e^{{\tan }^{-1}x}+C$
• D. $x\cdot e^{{\tan }^{-1}x}+C$

Answer 1

Answer: (C)

$I=\int e^{{\tan }^{-1}x}\left(1+x+x^2\right)\cdot \left(-\frac{1}{1+x^2}\right)dx$
$=-\int e^{{\tan }^{-1}x}\left(1+\frac{x}{1+x^2}\right)dx$
$=-\int e^{{\tan }^{-1}x}dx-\int x\cdot \frac{e^{{\tan }^{-1}x}}{1+x^2}dx$
$=-\int e^{{\tan }^{-1}x}dx-x\cdot e^{{\tan }^{-1}x}+\int e^{{\tan }^{-1}x}dx+C$
$=-x\cdot e^{{\tan }^{-1}x}+C$