If $\int \frac{x+1}{\sqrt{2x-1}} dx = f(x)\sqrt{2x-1} + C$, where $C$ is a constant of integration, then $f(x)$ is equal to:

JEE Main · 2019 · 11 Jan 2019 · Shift 2 · Indefinite Integrals

\int \frac{x + 1}{2 x - 1} d x = f (x) 2 x - 1 + C

, where

C

is a constant of integration, then

f (x)

is equal to:

Answer: (D)

Put

2 x - 1 = t^{2}

, so

d x = t d t

and

x = \frac{t ^{2} + 1}{2}

. Then

\int \frac{x + 1}{2 x - 1} d x = \int \frac{\frac{t ^{2} + 1}{2} + 1}{t} t d t = \int \frac{t ^{2} + 3}{2} d t = \frac{t ^{3}}{6} + \frac{3 t}{2} + C = t \cdot \frac{t ^{2} + 9}{6} + C

. Replacing

t = 2 x - 1

and

t^{2} = 2 x - 1

, this is

2 x - 1 \cdot \frac{2 x + 8}{6} + C = \frac{1}{3} (x + 4) 2 x - 1 + C

, so

f (x) = \frac{1}{3} (x + 4)

JEE Main · 2019 · 11 Jan 2019 · Shift 2 · Single correct option

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