CUET Preparation Today
Evaluate $\int \frac{(1 + \cos x)}{x + \sin x} dx$. |
$\ln|1 + \cos x| + C$ $\frac{1}{(x + \sin x)^2} + C$ $\ln|x + \sin x| + C$ $x + \sin x + C$ |
$\ln|x + \sin x| + C$ |
The correct answer is Option (3) → $\ln|x + \sin x| + C$ Consider that, $I = \int \frac{(1 + \cos x)}{(x + \sin x)} dx$ Let $x + \sin x = t \Rightarrow (1 + \cos x) dx = dt$ $∴I = \int \frac{1}{t} dt = \log |t| + C$ $= \log |(x + \sin x)| + C$ |
