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

$\frac{\sin x-x \cos x}{x \sin x+\cos x}+C$

We have,

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

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

$\Rightarrow I=x \sec x \times \frac{-1}{x \sin x+\cos x} -\int(\sec x+x \sec x \tan x) \times \frac{-1}{x \sin x+\cos x} d x$

$\Rightarrow I=\frac{-x \sec x}{x \sin x+\cos x}+\int \sec ^2 x d x$

$\Rightarrow I=\frac{-x \sec x}{x \sin x+\cos x}+\tan x+C=\frac{\sin x-x \cos x}{x \sin x+\cos x}+C$