Question

If $I=\int \frac{1}{e^x}{\tan }^{-1}\left(e^x\right)dx$, then I equals

• A. $-e^{-x}{\tan }^{-1}\left(e^x\right)+\log (1+e^{2x})+C$
• B. $x-e^{-x}{\tan }^{-1}{e}^x-\frac{1}{2}\log (1+e^x)+C$
• C. $x-e^{-x}{\tan }^{-1}\left(e^x\right)-\frac{1}{2}\log (1+e^{2x})+C$
• D. none of these

Answer 1

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

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

Put $e^x=t$

$I=\int \frac{1}{t^2}{\tan }^{-1}tdt$

We know that, $\int u.vdx=u\int vdx$ $-\int (\frac{du}{dx}\int vdx)dx$

$I=-\frac{1}{t}{\tan }^{-1}t+\int \frac{1}{t}\frac{1}{t^2+1}dt$

$I=-\frac{1}{t}{\tan }^{-1}t+\int [\frac{1}{t}-\frac{1}{t^2+1}]dt$

$I=-\frac{1}{e^x}{\tan }^{-1}\left(e^x\right)+x-\frac{1}{2}\log (e^{2x}+1)+c$

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