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JEE Main · 2026 · 28 Jan 2026 · Shift 2 · Indefinite Integrals

Let

f (x) = \int \frac{dx}{x ^{(\frac{2}{3})} + 2 x ^{(\frac{1}{2})}}

be such that

f (0) = - 26 + 24 lo g_{e} (2)

. If

f (1) = a + b lo g_{e} (3)

, where

a, b \in Z

, then

a + b

is equal to:

(A)-18
(B)-5
(C)-11✓ Correct
(D)-26

Answer: (C)

Solution

Substitute

x = t^{6}

, so

d x = 6 t^{5} d t

and

x^{2/3} = t^{4}, x^{1/2} = t^{3}

. Then

f (x) = \int \frac{6 t ^{5}}{t ^{4} + 2 t ^{3}} d t = 6 \int \frac{t ^{2}}{t + 2} d t = 6 \int (t - 2 + \frac{4}{t + 2}) d t = 3 t^{2} - 12 t + 24 ln (t + 2) + C .

Returning to

x

,

f (x) = 3 x^{1/3} - 12 x^{1/6} + 24 ln (x^{1/6} + 2) + C

. The condition

f (0) = 24 ln 2 + C = - 26 + 24 ln 2

gives

C = - 26

. Then

f (1) = 3 - 12 + 24 ln 3 - 26 = - 35 + 24 ln 3,

so

a = - 35, b = 24

and

a + b = - 11

.

JEE Main · 2026 · 28 Jan 2026 · Shift 2 · Single correct option

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