Question

Integrate $\frac{e^{{\tan }^{-1}}x}{1+x^2}$ with respect to $x$.

Answer 1

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

Put ${\tan }^{-1}x=t$

Diff wrt. $x$

$\frac{1}{1+x^2}dx=dt$.
$I=e^t+C$

$=e^{{\tan }^{-1}x}+C$