Let  $I\left(x\right)=\int \frac{x^2(x{\sec }^{2}x+\tan x)}{{(x\tan x+1)}^2}dx.$  If  $I\left(0\right)=0,$  then  $I\left(\frac{\pi }{4}\right)$  is equal to 

$$\begin{gathered} I\left(x\right)=x^2\left(\frac{-1}{x\tan x+1}\right)+\int \frac{2x}{x\tan x+1}dx \\ =\frac{-x^2}{x\tan x+1}+2\int \frac{x\cos x}{x\sin x+\cos x}dx \\ =\frac{-x^2}{x\tan x+1}+2\log (x\sin x+\cos x)+C \\ I\left(0\right)=0\Rightarrow C=0 \end{gathered}$$