$\int \frac{{\mathrm{x}}^2\left({\mathrm{xsec}}^2\mathrm{x}+\tan \mathrm{x}\right)}{(\mathrm{xtan}\mathrm{x}+1{)}^2}\mathrm{dx}=\frac{-{\mathrm{x}}^2}{\mathrm{xtan}\mathrm{x}+1}+\mathrm{f}\left(\mathrm{x}\right)+\mathrm{c}$ then $f\left(x\right) =$

$$\begin{gathered} \int \frac{{\mathrm{x}}^2\left({\mathrm{xsec}}^2\mathrm{x}+\mathrm{tanx}\right)}{(\mathrm{xtanx}+1{)}^2}\mathrm{dx} \\ \mathrm{GI}=x^2\left(\frac{-1}{x\tan x+1}\right)-\int \frac{2x(-1)}{x\tan x+1}dx \\ by u\sin g by parts \\ I=\frac{-x^2}{x\tan x+1}+2\int \frac{x\cos x}{x\sin x+\cos x}dx \\ \boxed{\frac{d}{dx}\left(x\sin x+\cos x\right)=x\cos x} \\ I=\frac{-x^2}{x\tan x+1}+2\log \left|x\sin x+\cos x\right|+C \end{gathered}$$