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

Divide the fraction into two parts by adding and subtracting x.

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

Now, Divide this into two parts like this.

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

Now Solve 1st term by letting $$\left(x\tan x+1\right)=u$$ as it is and for the 2nd term use $$\left(\log\left(x\tan x+1\right)\right)=F\left(x\right)$$ as we can see this in the options that log term is there so use this approach and double derivate this and put the boundary conditions as given in the question.

So,

Let $$\left(\log\left(x\tan x+1\right)\right)=F\left(x\right)$$ for the second part means,

$$I\left(x\right)=\log\left(x\tan x+1\right)\ -\ \frac{1}{x\tan x+1}+\ F\left(x\right)$$

Putting the I(0) = 0 in the above equation we will get F(x) = 0.

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

Your question will be solved.