$\int \frac{x \cos x}{(x \sin x+\cos x)^2} d x$ is equal to

$-\frac{1}{(x \sin x+\cos x)}+k$

Let $I=\int \frac{x \cos x}{(x \sin x+\cos x)^2} d x$

Let $\frac{1}{x \sin x+\cos x}= t \Rightarrow \frac{(x \sin x+\cos x) . 0-1(x \cos x+\sin x-\sin x)}{(x \sin x+\cos x)^2}=\frac{d t}{d x}$

$\Rightarrow \frac{-x \cos x}{(x \sin x+\cos x)^2}=\frac{d t}{d x}$

∴  $I=-\int d t=-\frac{1}{x \sin x+\cos x}+c$

Hence (2) is the correct answer.