Let I(x)=∫x2(xsec2x+tanx)(xtanx+1)2dx. If I(0)=0, then I(π4) is equal to

I(x)=x2(−1xtanx+1)+∫2xxtanx+1dx =−x2xtanx+1+2∫xcosxxsinx+cosxdx =−x2xtanx+1+2log(xsinx+cosx)+C I(0)=0⇒C=0

Let I(x)=∫x2(xsec2x+tanx)(xtanx+1)2dx. If I(0)=0, then I(π4) is equal to

I(x)=x2(−1xtanx+1)+∫2xxtanx+1dx =−x2xtanx+1+2∫xcosxxsinx+cosxdx =−x2xtanx+1+2log(xsinx+cosx)+C I(0)=0⇒C=0

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

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

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

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

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

answer is D.

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Detailed Solution

$I (x) = x^{2} (\frac{- 1}{x \tan x + 1}) + \int \frac{2 x}{x \tan x + 1} d x = \frac{- x^{2}}{x \tan x + 1} + 2 \int \frac{x \cos x}{x \sin x + \cos x} d x = \frac{- x^{2}}{x \tan x + 1} + 2 \log (x \sin x + \cos x) + C I (0) = 0 \Rightarrow C = 0$

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Let I(x)=∫x2(xsec2x+tanx)(xtanx+1)2dx. If I(0)=0, then I(π4) is equal to

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