Practicing Success
Practicing Success
CUET Preparation Today
$∫e^x\left(tan^{-1}x+\frac{1}{1+x^2}\right)dx $ is equal to : |
$e^x(\frac{1}{1+x^2})+C$ $tan^{-1}x+C$ $e^xtan^{-1}x+C$ $e^xcot^{-1}x+C$ |
$e^xtan^{-1}x+C$ |
The correct answer is Option (3) → $e^x\tan^{-1}x+C$ $∫e^x\left(\tan^{-1}x+\frac{1}{1+x^2}\right)dx$ $f(x)=\tan^{-1}x$, $f'(x)=\frac{1}{1+x^2}$ $=e^x\tan^{-1}x+C$ |
