Question

Integrate the function: $\frac{e^{\tan^{-1}}}x{1+x^2}$

Answer 1

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

∴$\frac{1}{1+x^2}dx=dt$

$\Rightarrow \int \frac{e^{\tan^{-1}}}{1+x^2}dx=\int e^tdt$

=$e^t+C$

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