$\int \frac{x^2\left(x{\sec }^{2}x+\tan x\right)}{(x\tan x+1{)}^2}dx$

$\int \underset{I}{\underbrace{x^2}}\cdot \underset{\mathrm{II}}{\underbrace{\frac{\left(x{\sec }^{2}x+\tan x\right)}{(x\tan x+1{)}^2}}}dx$

Integrating by parts 

$$\begin{gathered} I=x^2\cdot \left(\frac{-1}{x\tan x+1}\right)+2\int \frac{x}{x\tan x+1}dx \\ =\frac{-x^2}{x\tan x+1}+2\int \frac{x\cos x}{x\sin x+\cos x}dx \end{gathered}$$

Put $x\sin x+\cos x=t$

$$\begin{gathered} x\cos x\mathrm{dx}=\mathrm{dt} \\ =\frac{-x^2}{x\tan x+1}+2\int \frac{dt}{t}=\frac{-x^2}{x\tan x+1}+2\ln (x\sin x+\cos x)+C \end{gathered}$$