If the solve the problem

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 C is a constant of integration, then the function ƒ(x) is equal to- 

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

  2. $\frac{3}{x^{2}}$

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

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


Correct Option: , 4

Solution:

$\int \frac{d x}{x^{3}\left(1+x^{6}\right)^{2 / 3}}=x f(x)\left(1+x^{6}\right)^{1 / 3}+c$

$\int \frac{\mathrm{dx}}{\mathrm{x}^{7}\left(\frac{1}{\mathrm{x}^{6}}+1\right)^{2 / 3}}=\mathrm{x} f(\mathrm{x})\left(1+\mathrm{x}^{6}\right)^{1 / 3}+\mathrm{c}$'

Let $t=\frac{1}{x^{6}}+1$

$\mathrm{dt}=\frac{-6}{\mathrm{x}^{7}} \mathrm{dx}$

$-\frac{1}{6} \int \frac{\mathrm{dt}}{\mathrm{t}^{2 / 3}}=-\frac{1}{2} \mathrm{t}^{1 / 3}$

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

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

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