$∫e^x\left(tan^{-1}x+\frac{1}{1+x^2}\right)dx $ is equal to :
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → $e^x\tan^{-1}x+C$
$∫e^x\left(\tan^{-1}x+\frac{1}{1+x^2}\right)dx$
$f(x)=\tan^{-1}x$, $f'(x)=\frac{1}{1+x^2}$
$=e^x\tan^{-1}x+C$