If $\int \frac{d x}{x^{3}\left(1+x^{6}\right)^{2/3}}=x f(x)\left(1+x^{6}\right)^{\frac{1}{3}}+C$ where $C$ is a constant of integration, then the function $f(x)$ is equal to:

JEE Main · 2019 · 08 Apr 2019 · Shift 2 · Indefinite Integrals

\int \frac{d x}{x ^{3} ( 1 + x ^{6} ) ^{2/3}} = x f (x) (1 + x^{6})^{\frac{1}{3}} + C

where

C

is a constant of integration, then the function

f (x)

is equal to:

Answer: (B)

Pull out a factor so a clean substitution appears:

I = \int \frac{d x}{x ^{7} ( 1 + x ^{- 6} ) ^{2/3}}

. Let

t = 1 + x^{- 6}

, so

d t = - 6 x^{- 7} d x

, which means

\frac{d x}{x ^{7}} = - \frac{d t}{6}

. Then

I = - \frac{1}{6} \int t^{- 2/3} d t = - \frac{1}{6} \cdot 3 t^{1/3} + C = - \frac{1}{2} (1 + x^{- 6})^{1/3} + C = - \frac{( 1 + x ^{6} ) ^{1/3}}{2 x ^{2}} + C

. Matching this against

x f (x) (1 + x^{6})^{1/3} + C

gives

x f (x) = - \frac{1}{2 x ^{2}}

, so

f (x) = - \frac{1}{2 x ^{3}}

JEE Main · 2019 · 08 Apr 2019 · Shift 2 · Single correct option