Question

Find the integration of $\int \frac{e^{{\tan }^{-1}x}}{1+x^2}.dx$

• A. $e^{{\tan }^{-1}x}+C$
• B. $e^{-{\tan }^{-1}x}+C$
• C. $\frac{e^{{\tan }^{-1}x}}{2}+C$
• D. $e^{\tan x}+C$

Answer 1

The correct option is A $e^{{\tan }^{-1}x}+C$

Given, $\int \frac{e^{{\tan }^{-1}x}}{1+x^2}.dx$

Let $t={\tan }^{-1}x$
$⟹dt=\frac{1}{1+x^2}.dx$
$\Rightarrow dx=(1+x^2).dt$
Putting these value in above integration
$\therefore$ we have $I=\int e^t.dt=e^t+C$
$\therefore$ $I=e^{{\tan }^{-1}x}+C$