Find $\int e^x \left(\tan^{-1} x + \frac{1}{1+x^2}\right) dx$
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → $e^x \tan^{-1} x + C$
We have $I = \int e^x \left(\tan^{-1} x + \frac{1}{1+x^2}\right) dx$
Consider $f(x) = \tan^{-1} x$, then $f'(x) = \frac{1}{1+x^2}$.
Thus, the given integrand is of the form $e^x [f(x) + f'(x)]$.
Therefore, $I = \int e^x \left(\tan^{-1} x + \frac{1}{1+x^2}\right) dx = e^x \tan^{-1} x + C$.