Question:
Evaluate: $\int \frac{\mathrm{e}^{\tan ^{-1}}}{1+\mathrm{x}^{2}} \mathrm{dx}$
Solution:
Given, $\int \frac{e^{\tan ^{-1}}}{1+x^{2}} d x$
Let $\tan ^{-1} x=t$
$\partial \frac{d y}{d x}\left(\operatorname{Tan}^{-1} x\right)=\mathrm{dt}$
$\partial \frac{1}{1+x^{2}} d x=d t$
Now, $\int \frac{e^{\tan ^{-1}}}{1+x^{2}} d x$
$=\int e^{t} d t$
$=e^{t}+c$
$=e^{\tan ^{-1} x+c}$