Question

Solve:

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

• A. $e^x{\tan }^{-1}x+c$
• B. $\frac{e^x}{1+x^2}+c$
• C. $e^x\tan x+c$
• D. None of these

Answer 1

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

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

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

$\therefore I=e^x{\tan }^{-1}x+c$.