Let Ix=∫x2x sec2+tanx(x tanx+1)2dx If I0=0, then Iπ4 is equal to

Question

Let

I (x) = \int \frac{x^{2} (x \sec^{2} + \tan x)}{(x \tan x + 1)^{2}} d x

I (0) = 0,

then

I (\frac{π}{4})

is equal to

\log_{e} \frac{(π + 4)^{2}}{16} + \frac{π^{2}}{4 (π + 4)}

\log_{e} \frac{(π + 4)^{2}}{16} - \frac{π^{2}}{4 (π + 4)}

\log_{e} \frac{(π + 4)^{2}}{32} - \frac{π^{2}}{4 (π + 4)}

\log_{e} \frac{(π + 4)^{2}}{32} + \frac{π^{2}}{4 (π + 4)}

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