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)

Correct answer is (c) $-x\cdot e^{{\tan }^{-1}x}+c$