then the function f(x) is equal to :

Question:

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 $\mathrm{C}$ is a

constant of integration, then the function $f(x)$ is equal to :

1. (1) $\frac{3}{x^{2}}$

2. (2) $-\frac{1}{6 x^{3}}$

3. (3) $-\frac{1}{2 x^{2}}$

4. (4) $-\frac{1}{2 x^{3}}$

Correct Option: , 4

Solution:

Let, $\int \frac{d x}{x^{3}\left(1+x^{6}\right)^{\frac{2}{3}}}=\int \frac{d x}{x^{7}\left(1+x^{-6}\right)^{\frac{2}{3}}}$

Put $1+x^{-6}=t^{3} \quad \Rightarrow-6^{-7} d x=3 t^{2} d t \Rightarrow \frac{d x}{x^{7}}=\left(-\frac{1}{2}\right) t^{2} d t$

Now, $I=\int\left(-\frac{1}{2}\right) \frac{t^{2} d t}{t^{2}}=-\frac{1}{2} t+C$

$=-\frac{1}{2}\left(1+x^{-6}\right)^{\frac{1}{3}}+C=-\frac{1}{2} \frac{\left(1+x^{6}\right)^{\frac{1}{3}}}{x^{2}}+C$

$=-\frac{1}{2 x^{3}} x\left(1+x^{6}\right)^{\frac{1}{3}}+C$

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