If $f(\mathrm{x})=\int \frac{1}{\mathrm{x}^{1 / 4}\left(1+\mathrm{x}^{1 / 4}\right)} \mathrm{dx}, f(0)=-6$, then $f(1)$ is equal to :

(B) $4\left(\log _{\mathrm{e}} 2-2\right)$

(D) $4\left(\log _{e} 2+2\right)$

JEE Main · 2025 · 28 Jan 2025 · Shift 2 · Indefinite Integrals

f (x) = \int \frac{1}{x ^{1/4} ( 1 + x ^{1/4} )} dx, f (0) = - 6

, then

f (1)

is equal to :

Answer: (B)

Substitute

x = t^{4}

, so

d x = 4 t^{3} d t

and

x^{1/4} = t

. The integral becomes

\int \frac{4 t ^{3}}{t ( 1 + t )} d t = \int \frac{4 t ^{2}}{1 + t} d t

. Writing

4 t^{2} = 4 (t - 1) (t + 1) + 4

gives

\frac{4 t ^{2}}{1 + t} = 4 (t - 1) + \frac{4}{1 + t}

, so the integral is

2 t^{2} - 4 t + 4 ln (1 + t) + C

. Returning to

x

f (x) = 2 x^{1/2} - 4 x^{1/4} + 4 ln (1 + x^{1/4}) + C

. The condition

f (0) = - 6

gives

C = - 6

. Hence

f (1) = 2 - 4 + 4 ln 2 - 6 = 4 ln 2 - 8 = 4 (ln 2 - 2)

JEE Main · 2025 · 28 Jan 2025 · Shift 2 · Single correct option