Let I(x)=∫ (x²(x sec ² x+ tan x))/((x tan x+1)²) d x. If I(0)=0, then I((π)/4) is equal to:

(A) log e ((π+4)²)/32-(π²)/(4(π+4))

(B) log e ((π+4)²)/16-(π²)/(4(π+4))

(C) log e ((π+4)²)/16+(π²)/(4(π+4))

(D) log e ((π+4)²)/32+(π²)/(4(π+4))

Integral Calculus: JEE Main 2023 Mathematics question

JEE Main 2023 · 6 Apr, Morning shift · Q3997

JEE Main 20236 Apr, Morning shiftMathematicsMediumIntegral Calculus369× since 2002

Let

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

. If

I(0)=0

, then

I\left(\frac{\pi}{4}\right)

is equal to :

JEE Main 2023 · 6 Apr, Morning shift · Q3997

JEE Main 20236 Apr, Morning shiftMathematicsMediumIntegral Calculus369× since 2002

Let

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

. If

I(0)=0

, then

I\left(\frac{\pi}{4}\right)

is equal to :