Question

Evaluate $\int e^x({\tan }^{-1}x+\frac{1}{1+x^2})dx$
(where $C$ is constant of integration)

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

Answer 1

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

Let $I=\int e^x({\tan }^{-1}x+\frac{1}{1+x^2})dx$
This is of the form $\int e^x\left(f\right(x)+f^{\prime }(x\left)\right)dx=e^{x}f\left(x\right)$
Where $f\left(x\right)={\tan }^{-1}x,f^{\prime }\left(x\right)=\frac{1}{1+x^2}$
$\therefore I=e^x{\tan }^{-1}x+C$