CUET Preparation Today
Find $\int e^x \left(\tan^{-1} x + \frac{1}{1+x^2}\right) dx$ |
$e^x \left(\frac{1}{1+x^2}\right) + C$ $e^x \tan^{-1} x + C$ $\frac{e^x}{1+x^2} + \tan^{-1} x + C$ $e^x (\tan^{-1} x)^2 + C$ |
$e^x \tan^{-1} x + C$ |
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, $I = \int e^x \left(\tan^{-1} x + \frac{1}{1+x^2}\right) dx = e^x \tan^{-1} x + C$. |
