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

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

Apply Integration by parts.

$\int f\left(x\right)g\left(x\right)dx=f\left(x\right)\int g\left(x\right)dx-\int \left(f\text{'}\left(x\right)\int g\left(x\right)dx\right)dx$

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

Let $x\tan x+1=p$

$\Rightarrow \left(x {\sec }^{2}x+\tan x\right)dp=dx$

$=x^2\int \frac{dp}{p^2}-\int \left(2x\int \frac{dp}{p^2}\right)dx$

$=\frac{-x^2}{(x \tan x+1)}+\int \frac{2x}{x \tan x+1}dx$

Let $I_1=2\int \frac{x}{x \tan x+1}dx$

$=2\int \frac{x \cos x}{x \sin x+\cos x}dx$

Let $x \sin x+\cos x=t$

$\Rightarrow (x \cos x+\sin x-\sin x)dx=dt$

$\Rightarrow (x \cos x)dx=dt$

$=2\int \frac{dt}{t}=2\log t+c$

$=2\log |x \sin x+\cos x|+c$

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

$=\frac{-x^2}{x \tan x+1}+2\log |x \sin x+\cos x|+c$

But $I\left(0\right)=0$

$\Rightarrow c=0$

Also,$I\left(\frac{\pi }{4}\right)=-\frac{{\left(\frac{\pi }{4}\right)}^2}{\frac{\pi }{4}\times 1+1}+2\log \left|\frac{1}{\sqrt{2}}\left(\frac{\pi }{4}\right)+\frac{1}{\sqrt{2}}\right|$

$={\log }_e\frac{(\pi +4{)}^2}{32}-\frac{{\pi }^2}{4(\pi +4)}$

Hence, this is the correct option.