Question

$\int \frac{{\cot }^{-1}\left(e^x\right)}{e^x}dx$ is equal to

• A. $\frac{1}{2}\ln (e^{2x}+1)-\frac{\cot \left(e^x\right)}{e^x}+x+c$
• B. $\frac{1}{2}\ln (e^{2x}+1)+\frac{{\cot }^{-1}\left(e^x\right)}{e^x}+x+c$
• C. $\frac{1}{2}\ln (e^{2x}+1)-\frac{{\cot }^{-1}\left(e^x\right)}{e^x}-x+c$
• D. $\frac{1}{2}\ln (e^{2x}+1)+\frac{{\cot }^{-1}\left(e^x\right)}{e^x}-x+c$

Answer 1

The correct option is C $\frac{1}{2}\ln (e^{2x}+1)-\frac{{\cot }^{-1}\left(e^x\right)}{e^x}-x+c$

$Let{\cot }^{-1}\left(e^x\right)=\theta \Rightarrow e^x=\cot \theta$
$\therefore e^{x}dx=-cose{c}^2\theta d\theta$
$\int \frac{\theta }{cot\theta }\times \left(\frac{-cose{c}^2\theta d\theta }{cot\theta }\right)$
$=-\int \theta se{c}^2\theta d\theta =-\left[\right(\theta \tan \theta )-\int 1.\tan \theta d\theta ]$
$=-\theta \tan \theta +\ln \left(\sec \theta \right)+c$



$=\ln (\frac{\sqrt{1+e^{2x}}}{e^x})-\frac{{\cot }^{-1}\left(e^x\right)}{e^x}+c$
$=\frac{1}{2}\ln (e^{2x}+1)-\frac{{\cot }^{-1}\left(e^x\right)}{e^x}-x+c$