Question

$\int e^{{\tan }^{-1}x}\left[\frac{1}{1+x^2}\right]dx=$

• A. $x^2{e}^{{\tan }^{-1}x}+c$
• B. $x{e}^{{\tan }^{-1}x}+c$
• C. $e^{{\tan }^{-1}x}+c$
• D. $\frac{1}{2}{e}^{{\tan }^{-1}x}+c$

Answer 1

Answer: (C)

$e^{{\tan }^{-1}x}+c$

We have

$\frac{d}{dx}\left({\tan }^{-1}x\right)=\frac{1}{1+x^2}$

Now,

$\int e^{{\tan }^{-1}x}\left[\frac{1}{1+x^2}\right]dx$

$=\int e^{{\tan }^{-1}x}d\left[{\tan }^{-1}x\right]$

$=e^{{\tan }^{-1}x}+c$ [ Where $c$ is integrating constant]