Let I (x) = ∫ x ^ 2 (x sec ^ 2 + x) ( xtanx + 1 ) ^ 2 dx If I(0) =...

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

Apply Integration by parts.

$∫ f\left(x\right)g\left(x\right)dx = f\left(x\right) ∫ g\left(x\right)dx - ∫ \left(f ' \left(x\right) ∫ g\left(x\right)dx\right)dx$

$= {x}^{2} ∫ \left(\dfrac{x{sec}^{2}x + \tan x}{( xtanx + 1{)}^{2}}\right)dx - ∫ \left(\dfrac{d {x}^{2}}{d x} ∫ \left(\dfrac{x{sec}^{2}x + \tan x}{( xtanx + 1{)}^{2}}\right)dx\right)dx$

Let $xtanx + 1 = p$

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

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

$= \dfrac{- {x}^{2}}{( xtanx + 1 )} + ∫ \dfrac{2x}{xtanx + 1}dx$

Let ${I}_{1} = 2 ∫ \dfrac{x}{xtanx + 1}dx$

$= 2 ∫ \dfrac{xcosx}{xsinx + \cos x}dx$

Let $xsinx + \cos x = t$

$⇒ ( xcosx + \sin x - \sin x ) dx = dt$

$⇒ ( xcosx ) dx = dt$

$= 2 ∫ \dfrac{dt}{t} = 2\log t + c$

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

$∴ ∫ \dfrac{{x}^{2}\left(x{sec}^{2}x + \tan x\right)}{( xtanx + 1{)}^{2}}dx$

$= \dfrac{- {x}^{2}}{xtanx + 1} + 2log | xsinx + \cos x | + c$

But $I ( 0 ) = 0$

$⇒ c = 0$

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

$= \log_{e}\dfrac{( \pi + 4{)}^{2}}{32} - \dfrac{{\pi}^{2}}{4 ( \pi + 4 )}$

Hence, this is the correct option.