Let I(x)=∫ 3 d x (4 x+6)( 4 x^ 2 +8 x+3 ) and I(0)= 3 4 +20 . If I((1)/(2))= a 2 b +c , where a, b, c ∈ N, gcd(a, b)=1 , then a+b+c is equal to

Correct answer: B. 31 4x^2+8x+3 = (2x+1)(2x+3) and 4x+6 = 2(2x+3) . I(x) = (3)/(4)∫ du u^ 3/2 u-2 where u = 2x+3 . Substituting u = 2 ^2φ : I(x) = (3)/(4) φ + C = (3)/(4) (2x+1)/(2x+3) + C . I(0) = (3)/(4) (1)/(3) + C = 3 4 + C = 3 4 + 20 C = 20 . I((1)/(2)) = (3)/(4) (2)/(4) + 20 = 3 4 2 + 20 = 3 2 8 + 20 . a = 3, b = 8, c = 20 , (3,8) = 1 . a+b+c = 31 .

JEE Main 202623 JanuaryEvening Shift

Exam: JEE Main
Year: 2026
Date: 23 January
Shift: Evening Shift
Subject: Mathematics
Chapter: Indefinite Integration

JEE Main 2026 · 23 January · Evening Shift · Mathematics · Indefinite Integration

Let I(x)=∫ 3 d x (4 x+6)( 4 x^ 2 +8 x+3 ) and I(0)= 3 4 +20 . If...

Let

\mathrm{I}(x)=\int \frac{3 d x}{(4 x+6)\left(\sqrt{4 x^{2}+8 x+3}\right)} and

\mathrm{I}(0)=\frac{\sqrt{3}}{4}+20

. If

\mathrm{I}\left(\frac{1}{2}\right)=\frac{a \sqrt{2}}{b}+\mathrm{c}

, where

a, b, \mathrm{c} \in \mathrm{N}, \operatorname{gcd}(a, b)=1

, then

a+b+c is equal to

Options

Correct answer

B. 31

Step-by-step solution

4x^2+8x+3 = (2x+1)(2x+3)

and

4x+6 = 2(2x+3)

I(x) = \frac{3}{4}\int \frac{du}{u^{3/2}\sqrt{u-2}}

where

u = 2x+3

.
Substituting

u = 2\sec^2\phi

I(x) = \frac{3}{4}\sin\phi + C = \frac{3}{4}\sqrt{\frac{2x+1}{2x+3}} + C

I(0) = \frac{3}{4}\sqrt{\frac{1}{3}} + C = \frac{\sqrt{3}}{4} + C = \frac{\sqrt{3}}{4} + 20 \Rightarrow C = 20

I\left(\frac{1}{2}\right) = \frac{3}{4}\sqrt{\frac{2}{4}} + 20 = \frac{3}{4\sqrt{2}} + 20 = \frac{3\sqrt{2}}{8} + 20

a = 3, b = 8, c = 20

\gcd(3,8) = 1

a+b+c = 31

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