Question

If $\int \frac{dx}{x^3(1+x^6{)}^{2/3}}=x$. $f\left(x\right)\cdot (1+x^6{)}^{1/3}+C$, then $f\left(x\right)$ is equal to?

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

Answer 1

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

$\int \frac{dx}{x^3(1+x^6{)}^{2/3}}\Rightarrow \int \frac{dx}{x^7{(1+\frac{1}{x^6})}^{2/3}}\Rightarrow {(1+\frac{1}{x^6})}^{1/3}=t$
$-\frac{6}{x^7}dx=dt$
$\Rightarrow -\frac{1}{6}\int \frac{dt}{t^{2/3}}$
$-\frac{1}{6}\left(\frac{t^{1/3}}{\frac{1}{3}}\right)=-\frac{1}{2}\left[{(1+\frac{1}{x^6})}^{1/3}\right]+C=-\frac{1}{2}\frac{(1+x^6{)}^{1/3}}{x^2}+C=x$ $f\left(x\right)(1+x^6{)}^{1/3}$
$f\left(x\right)=-\frac{1}{2x^3}$.