Substitue $e^{x}=t$ and $e^{x} d x=d t$
Therefore
$I=\int \frac{\cot ^{-1}(t)}{t^{2}} d t$
Integrating it by part, taking $\cot ^{-1}(t)$ as first function and $t^{2}$ as second function, we get $I=-\frac{\cot ^{-1}(t)}{t}-\int \frac{1}{t\left(t^{2}+1\right)} d t$
$\int \frac{1}{t\left(t^{2}+1\right)} d t=P$
$t^{2}=u, \quad 2 t d t=d u$
$P=\frac{1}{2} \int \frac{1}{u(u+1)} d u$
$=\frac{1}{2} \int\left(\frac{1}{u}-\frac{1}{(u+1)}\right) d u$
$=\frac{1}{2}(\log u-\log u+1)+C$
$I=-\frac{\cot ^{-1}(t)}{t}-\frac{1}{2}(\log |u|-\log |u+1|)+C$
$=\frac{1}{2}\left(\log \left|t^{2}+1\right|-\log \left|t^{2}\right|-\frac{2 \cot ^{-1}(t)}{t}\right)+C$
$=\frac{1}{2}\left(\log \left|e^{2 x}+1\right|-\log \left|e^{2 x}\right|-\frac{2 \cot ^{-1}\left(e^{x}\right)}{e^{x}}\right)+C$
$=\frac{1}{2} \log \left|e^{2 x}+1\right|-\frac{\cot ^{-1}\left(e^{x}\right)}{e^{x}}-x+C$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$1.$ If $1$ ball is drawn from each of the boxes $B_1, B_2$ and $B_3$, the probability that all $3$ drawn balls are of the same colour is
$(A)$ $\frac{82}{648}$ $(B)$ $\frac{90}{648}$ $(C)$ $\frac{558}{648}$ $(D)$ $\frac{566}{648}$
$2.$ If $2$ balls are drawn (without replacement) from a randomly selected box and one of the balls is white and the other ball is red, the probability that these $2$ balls are drawn from bo $B _2$ is
$(A)$ $\frac{116}{181}$ $(B)$ $\frac{126}{181}$ $(C)$ $\frac{65}{181}$ $(D)$ $\frac{55}{181}$
Give the answer question $1$ and $2.$