Question

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

• A. $-\frac{1}{6x^3}$
• B. $-\frac{1}{2x^2}$
• C. $-\frac{1}{2x^3}$
• D. $\frac{3}{x^2}$

Answer 1

The correct option is C $-\frac{1}{2x^3}$

$I=\int \frac{dx}{x^3(1+x^6{)}^{2/3}}$

$=\int \frac{dx}{x^7{(1+\frac{1}{x^6})}^{2/3}}$

Let $1+\frac{1}{x^6}=t\Rightarrow \frac{-6}{x^7}dx=dt$
$\therefore I=-\frac{1}{6}\int \frac{dt}{t^{2/3}}$
$=-\frac{3}{6}{t}^{1/3}+C$
$=-\frac{1}{2}{(1+\frac{1}{x^6})}^{1/3}+C$
$=-\frac{1}{2x^2}(1+x^6{)}^{1/3}+C$

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