CUET Preparation Today
The value of the integral $I=\int e^x\left(\tan ^{-1} x+\frac{1}{1+x^2}\right) d x$ is: |
$\frac{e^x}{1+x^2}+C$, where C is a constant $e^x \tan ^{-1} x+C$, where C is a constant $\frac{1}{1+x^2}+C$, where C is a constant $\tan ^{-1} x+C$, where C is a constant |
$e^x \tan ^{-1} x+C$, where C is a constant |
The correct answer is Option (2) - $e^x \tan ^{-1} x+C$, where C is a constant |
