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

$\begin{array}{l}\text{ Let I=}\int \frac{dx}{x^3{\left(1+x^6\right)}^{\frac{2}{3}}}= {\int }_{ }^{ }\frac{dx}{x^3{\left(x^6{\left({\displaystyle \frac{1}{x^6}}+1\right)}^{ }\right)}^{{\displaystyle \raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}}=\int \frac{dx}{x^7{\left(1+x^{-6}\right)}^{\frac{2}{3}}}\\ \text{ Put }1+x^{-6}=t^3\Rightarrow -6x^{-7}dx=3t^{2}dt\\ \Rightarrow \frac{dx}{x^7}=\left(-\frac{1}{2}\right)t^{2}dt\\ \text{ Now, }I=\int \left(-\frac{1}{2}\right)\frac{t^{2}dt}{t^2}=-\frac{1}{2}t+C\\ =-\frac{1}{2}{\left(1+x^{-6}\right)}^{\frac{1}{3}}+C=-\frac{1}{2}\frac{{\left(1+x^6\right)}^{\frac{1}{3}}}{x^2}+C\\ =-\frac{1}{2x^3}x{\left(1+x^6\right)}^{\frac{1}{3}}+C\\ \text{ Hence, }f\left(x\right)=-\frac{1}{2x^3}\\ \Rightarrow \frac{-C}{2x^3}=\frac{-1}{2x^3}\\ \Rightarrow C=1\end{array}$