If $\int \frac{x^2(x{\sec }^{2}x+\tan x)}{{(x\tan x+1)}^2}dx=\frac{-x^2}{x\tan x+1}+f\left(x\right)+C$ 
Then f(x)=
 

$\begin{array}{l}I=\int \frac{x^2\left(x{\sec }^{2}x+\tan x\right)}{(x\tan x+1{)}^2}dx\\ =\int x^2\times \frac{\left(x{\sec }^{2}x+\tan x\right)}{(x\tan x+1{)}^2}dx\\ =\frac{-x^2}{(x\tan x+1)}+\int \frac{2x}{(x\tan x+1)}dx\\ I_1=\int \frac{2x}{(x\tan x+1)}dx\\ =2\int \frac{x\cos x}{(x\sin x+\cos x)}dx\end{array}$
Let $x\sin x+\cos x=t$
$x\cos xdx=dt$
$\begin{array}{l}\therefore I_1=\int \frac{2dt}{t}=2\log t\\ I_1=2\log |x\sin x+\cos x|\\ I=\frac{-x^2}{x\tan x+1}+2\log |x\sin x+\cos x|+C\\ \therefore f\left(x\right)=2\log |x\sin x+\cos x|\end{array}$