If $\int \frac{x^{2}dx}{{(x\sin x+\cos x)}^2}=\frac{-x}{\cos x(x\sin x+\cos x)}+f\left(x\right)+c$ where c is a constant. Then $f\left(x\right)$ is equal to 

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

Multiplying and dividing it by $\left(x\cos x\right),$ we get $\begin{array}{l}I=\int \left(x\sec x\right).\frac{\left(x\cos x\right)}{{(x\sin x+\cos x)}^2}dx\\ III\end{array}$

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

$=x\sec x.\frac{-1}{(x\sin x+\cos x)}-\int (x\sec x.\tan x+\sec x).\frac{-1}{x\sin x+\cos x}dx$

$=\frac{-x\sec x}{(x\sin x+\cos x)}+\int \frac{(x\sin x+\cos x)}{{\cos }^{2}x.(x\sin x+\cos x)}dx=-\frac{x\sec x}{x\sin x+\cos x}+\tan x+c$