$\int \frac{x^{2}dx}{{\left(x\sin x+\cos x\right)}^2}=\int \frac{x}{\cos x}\cdot \frac{x\cos x}{{\left(x\sin x+\cos x\right)}^2}dx$

$=\frac{x}{\cos x}\int \frac{x\cos x}{{\left(x\sin x+\cos x\right)}^2}dx-\int \left[\frac{d}{dx}\left(x\sec x\right)\int \frac{x\cos x}{{\left(x\sin x+\cos x\right)}^2}dx\right]dx$

$=\frac{x}{\cos x}\left(-\frac{1}{x\sin x+\cos x}\right)+\int {\sec }^{2}xdx$, $\text{∵\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\int \frac{x\cos x}{{\left(x\sin x+\cos x\right)}^2}dx=\frac{-1}{x\sin x+\cos x}$

$\& \frac{d}{dx}\left(x\sec x\right)=\sec x+x\sec x\tan x=\sec x\left(1+\frac{x\sin x}{\cos x}\right)={\sec }^{2}x\left(x\sin x+\cos x\right)$

$=\frac{-x}{\cos x\left(x\sin x+\cos x\right)}+\frac{\sin x}{\cos x}+C$

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