Let $\mathrm{I}(\mathrm{x})=\int \frac{\mathrm{x}^{2}(\mathrm{xsec}^{2} \mathrm{x}+\tan \mathrm{x})}{(\mathrm{x} \tan \mathrm{x}+1)^{2}} \mathrm{dx}$. If $\mathrm{I}(0)=0$ the I $\left(\frac{\pi}{4}\right)$ is equal to

(C) $\log _{e} \frac{(\pi+4)^{2}}{32}-\frac{\pi^{2}}{4(\pi+4)}$

(A) $\log _{\mathrm{e}} \frac{(\pi+4)^{2}}{16}-\frac{\pi^{2}}{4(\pi+4)}$

(B) $\log _{\mathrm{e}} \frac{(\pi+4)^{2}}{16}+\frac{\pi^{2}}{4(\pi+4)}$

(D) $\log _{e} \frac{(\pi+4)^{2}}{32}+\frac{\pi^{2}}{4(\pi+4)}$

Indefinite Integrals — JEE Main · 2023 · 06 Apr 2023 · Shift 1 Question with Solution

Let

I (x) = \int \frac{x ^{2} ( xsec ^{2} x + t a n x )}{( x t a n x + 1 ) ^{2}} dx

. If

I (0) = 0

the I

(\frac{π}{4})

is equal to

Answer: (C)

Let

u = x tan x + 1

, so

d u = (x sec^{2} x + tan x) d x

. The integral becomes

\int \frac{x ^{2}}{u ^{2}} d u

. Using integration by parts with

d v = u^{- 2} d u

and

v = - u^{- 1}

I (x) = - \frac{x ^{2}}{x t a n x + 1} + \int \frac{2 x}{x t a n x + 1} d x

. The remaining integral can be written as

2 \int \frac{x c o s x}{x s i n x + c o s x} d x = 2 ln ∣ x sin x + cos x ∣ + C

. Using

I (0) = 0

0 = - 0 + 2 ln 1 + C

, so

C = 0

. At

x = π /4

I (π /4) = - \frac{( π /4 ) ^{2}}{( π /4 ) \cdot 1 + 1} + 2 ln (\frac{π}{4} \cdot \frac{1}{2} + \frac{1}{2}) = - \frac{π ^{2} /16}{( π + 4 ) /4} + 2 ln \frac{π + 4}{4 2} = - \frac{π ^{2}}{4 ( π + 4 )} + ln \frac{( π + 4 ) ^{2}}{32}

JEE Main · 2023 · 06 Apr 2023 · Shift 1 · Single correct option