If ∫ dx x^3 (1+x^6)^ 2 3 =xf(x)(1+x^6)^ 1 3+C, where C is a constant of integration, then the function f(x) is equal to

We have, $\displaystyle \int \frac{dx}{{x}^{3}{(1+{x}^{6})}^{\frac{2}{3}}}=xf(x){(1+{x}^{6})}^{\frac{1}{3}}+C$∫x3(1+x6)32​dx​=xf(x)(1+x6)31​+C

Now, $\displaystyle I=\int \frac{dx}{{x}^{7}{(\frac{1}{{x}^{6}}+1)}^{\frac{2}{3}}}$I=∫x7(x61​+1)32​dx​

Put, $t=\frac{1}{{x}^{6}}+1$t=x61​+1$\Rightarrow dt=-\frac{6}{{x}^{7}}dx$⇒dt=−x76​dx

$\displaystyle \therefore I=-\frac{1}{6}\int \frac{dt}{{t}^{\frac{2}{3}}}=-\frac{1}{2}{t}^{\frac{1}{3}}+C$∴I=−61​∫t32​dt​=−21​t31​+C, where $C$C is the constant of integration$=-\frac{1}{2}{(\frac{1}{{x}^{6}}+1)}^{\frac{1}{3}}+C=-\frac{1}{2}\frac{{(1+{x}^{6})}^{\frac{1}{3}}}{{x}^{2}}+C$=−21​(x61​+1)31​+C=−21​x2(1+x6)31​​+C

$$\therefore f(x)=-\frac{1}{2{x}^{3}}$$∴f(x)=−2x31​