Solve the following equations:

Question:

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

where $\mathrm{C}$ is a constant of integration, then the function $f(\mathrm{x})$ is equal to-

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

  2.  $\frac{3}{\mathrm{x}^{2}}$

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

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


Correct Option: , 4

Solution:

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

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

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

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

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

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

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

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