Question

Evaluate $\int e^x({\tan }^{-1}x+\frac{1}{1+x^2})dx$.

Answer 1

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

As we know that,

$\int e^x[f\left(x\right)+f^{\prime }\left(x\right)]dx=e^{x}f\left(x\right)+c$

$\because \frac{d}{dx}\left({\tan }^{-1}x\right)=\frac{1}{1+x^2}$

$\therefore \int e^x({\tan }^{-1}x+\frac{1}{1+x^2})dx=e^x\left({\tan }^{-1}x\right)+c$