Question

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

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

Answer 1

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

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

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