Question

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

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

Answer 1

The correct option is C $-x{e}^{{\tan }^{-1}x}+c$

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