The value of the integral $I=\int e^x\left(\tan ^{-1} x+\frac{1}{1+x^2}\right) d x$ is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) - $e^x \tan ^{-1} x+C$, where C is a constant
$\int e^x\left(\underbrace{\tan ^{-1} x}_{f(x)}+\underbrace{\frac{1}{1+x^2}}_{f'(x)}\right) d x$
$=e^x \tan ^{-1} x+C$