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JEE Main · 2026 · 23 Jan 2026 · Shift 2 · Indefinite Integrals

Let

I (x) = \int \frac{3 dx}{( 4 x + 6 ) ( 4 x ^{2} + 8 x + 3 )}

and

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

. If

I (\frac{1}{2}) = \frac{a 2}{b} + c

, where

a, b, c \in N, gcd (a, b) = 1

, then

a + b + c

is equal to :

(A)29
(B)28
(C)31✓ Correct
(D)30

Answer: (C)

Solution

With

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

and

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

, differentiating

\frac{3}{4} \frac{2 x + 1}{2 x + 3}

reproduces the integrand, so

I (x) = \frac{3}{4} \frac{2 x + 1}{2 x + 3} + c

. At

x = 0

,

I (0) = \frac{3}{4} \cdot \frac{1}{3} + c = \frac{3}{4} + c

; matching

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

gives

c = 20

. Then

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

, so

a = 3

,

b = 8

,

c = 20

and

a + b + c = 31

.

JEE Main · 2026 · 23 Jan 2026 · Shift 2 · Single correct option

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