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

Since, \(\frac{d}{d x}(x \sin x+\cos x)=x \cos x\)
\(\therefore I=\int \frac{x^{2} d x}{(x \sin x+\cos x)^{2}}\)
\(=\int \frac{x}{\cos x} \cdot \frac{x \cos x}{(x \sin x+\cos x)^{2}} d x\)
\(=\frac{x}{\cos x} \cdot\left(\frac{-1}{x \sin x+\cos x}\right) -\int \frac{\cos x-x(-\sin x)}{\cos ^{2} x}\) \(\cdot \frac{-1}{(x \sin x+\cos x)} d x \)
\( =\frac{-x}{\cos x(x \sin x+\cos x)}+\int \sec ^{2} d x \)
\( =\frac{-x}{\cos x(x \sin x+\cos x)}+\tan x+C \)
\( =\frac{-x+\sin x(x \sin x+\cos x)}{\cos x(x \sin x+\cos x)}+C \)
\( =\frac{-x \cos ^{2} x+\sin x \cdot \cos x}{\cos x(x \sin x+\cos x)}+C \)
\( =\left(\frac{\sin x-x \cos x}{\cos x+x \sin x}\right)+C\)