Substitue $e ^{ x }= t$ and $e ^{ x } dx = dt$
Therefore
$I =\int \frac{\cot ^{-1}( t )}{ t ^{2}} dt$
Integrating it by part, taking $\cot ^{-1}( t )$ as first function and $t ^{2}$ as second function, we get $I =-\frac{\cot ^{-1}( t )}{ t }-\int \frac{1}{ t \left( t ^{2}+1\right)} dt$
$\int \frac{1}{t\left(t^{2}+1\right)} d t=P$
$t ^{2}= u , \quad 2 tdt = du$
$P =\frac{1}{2} \int \frac{1}{ u ( u +1)} du$
$=\frac{1}{2} \int\left(\frac{1}{ u }-\frac{1}{( u +1)}\right) du$
$=\frac{1}{2}(\log u -\log u +1)+ C $
$ I =-\frac{\cot ^{-1}( t )}{ t }-\frac{1}{2}(\log | u |-\log | u +1|)+ C $
$=\frac{1}{2}\left(\log \left| t ^{2}+1\right|-\log \left| t ^{2}\right|-\frac{2 \cot ^{-1}( t )}{ t }\right)+ C$
$=\frac{1}{2}\left(\log \left| e ^{2 x }+1\right|-\log \left| e ^{2 x }\right|-\frac{2 \cot ^{-1}\left( e ^{ x }\right)}{ e ^{ x }}\right)+ C $
$=\frac{1}{2} \log \left| e ^{2 x }+1\right|-\frac{\cot ^{-1}\left( e ^{ x }\right)}{ e ^{ x }}- x + C $
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.