Question

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

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

Answer 1

Answer: (C)

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

Let $\int \frac{{\cot }^{-1}\left(e^x\right)}{e^x}dx=I$
Substitue $e^x=t$ and $e^{x}dx=dt$
Therefore
$I=\int \frac{{\cot }^{-1}\left(t\right)}{t^2}dt$
Integrating it by part, taking ${\cot }^{-1}\left(t\right)$ as first function and $t^2$ as second function, we get
$I=-\frac{{\cot }^{-1}\left(t\right)}{t}-\int \frac{1}{t(t^2+1)}dt\int \frac{1}{t(t^2+1)}dt=P{t}^2=u,2tdt=duP=\frac{1}{2}\int \frac{1}{u(u+1)}du=\frac{1}{2}\int (\frac{1}{u}-\frac{1}{(u+1)})du=\frac{1}{2}(\log u-\log u+1)+CI=-\frac{{\cot }^{-1}\left(t\right)}{t}-\frac{1}{2}(\log \left|u\right|-\log |u+1|)+C=\frac{1}{2}(\log |t^2+1|-\log \left|t^2\right|-\frac{{2\cot }^{-1}\left(t\right)}{t})+C=\frac{1}{2}(\log |e^{2x}+1|-\log \left|e^{2x}\right|-\frac{{2\cot }^{-1}\left(e^x\right)}{e^x})+C=\frac{1}{2}\log |e^{2x}+1|-\frac{{\cot }^{-1}\left(e^x\right)}{e^x}-x+C$