Question

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

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

Answer 1

Answer: (D)

$\int \frac{dx}{x^3{\left(1+x^6\right)}^{\frac{2}{3}}}=xf\left(x\right){\left(1+x^6\right)}^{\frac{1}{3}}+c$
$\int \frac{dx}{x^7{\left(\frac{1}{x^6}+1\right)}^{\frac{2}{3}}}=xf\left(x\right){\left(1+x^6\right)}^{\frac{1}{3}}+c$
Let $t=\frac{1}{x^6}+1$
$dt=\frac{-6}{x^7}dx$
$=-\frac{1}{6}\int \frac{dt}{t^{\frac{2}{3}}}=-\frac{1}{2}{t}^{\frac{1}{3}}$
$=-\frac{1}{2}{\left(\frac{1}{x^6}+1\right)}^{\frac{1}{3}}=-\frac{1}{2}\frac{{\left(1+x^6\right)}^{\frac{1}{3}}}{x^2}$
$\therefore f\left(x\right)=-\frac{1}{2x^3}$