Question

Solve $\int \frac{e^{{\tan }^{-1}x}}{1+x^2}dx$

Answer 1

$Let,I=\int \frac{e^{{\tan }^{-1}}x}{1+x^2}dx$

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

$⟹\frac{1}{1+x^2}dx=dt$

$\therefore I=\int e^{t}dt$

$I=e^t+c=e^{{\tan }^{-1}x}+c$ (Ans)