Question

Evaluate $\int \frac{e^{{\tan }^{-1}x}}{1+x^2}dx,x\in \left(0,\infty \right)$

Answer 1

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