Evaluate $ \int\left(x^{3} + x^{6}\right)\left(x^{3} + 2\right)^{1/3} \,\mathrm{d} x $

Question

Integrate : $$ \int\left(x^{3} + x^{6}\right)\left(x^{3} + 2\right)^{1/3} \,\mathrm{d} x $$ I have tried assuming that ${{x}^{3}}+2$ term as ${u}^{3}$, $dx=u^2du/(u^3-2)^{2/3}$ $$ \int\left(x^{3} + x^{6}\right)u \,\mathrm{d} x=\int (u^3-2)(u^3-1)u \,\frac{u^2du}{(u^3-2)^{2/3}} $$ $$ =\int \,\frac{u^3(u^6-3u^3+2) du}{(u^3-2)^{2/3}} $$

but that does not take me anywhere. Please provide an alternate approach or help me how to proceed after this step.

Answer 1

Let $t=1+x^3$: $$I=\frac{1}{3} \int t {\left(t-1\right)}^{\frac{1}{3}} {\left(t+1\right)}^{\frac{1}{3}} \; dt$$ $$I=\frac{1}{3} \int t {\left(t^2-1\right)}^{\frac{1}{3}} \; dt$$ Finally, let $\xi=t^2-1$: $$I=\frac{1}{6} \int {\xi}^{\frac{1}{3}} \; d\xi$$ $$I=\frac{{\xi}^{\frac{4}{3}}}{8}+C$$ Substitute back to $t$: $$I=\frac{{\left(t^2-1\right)}^{\frac{4}{3}}}{8}+C$$ Substitute back to $x$: $$I=\frac{{\left(x^6+2x^3\right)}^{\frac{4}{3}}}{8}+C$$ $$\boxed{I=\frac{x^4 {\left(x^3+2\right)}^{\frac{4}{3}}}{8}+C}$$

Answer 2

You can avoid substitution as follows $$\int (x^3+x^6)(x^3+2)^{1/3}\ dx=\int x(x^2+x^5)(x^3+2)^{1/3}\ dx$$ $$=\int (x^5+x^2)(x^6+2x^3)^{1/3}\ dx$$ $$=\frac16\int 6(x^5+x^2)(x^6+2x^3)^{1/3}\ dx$$ $$=\frac16\int (x^6+2x^3)^{1/3}\ d(x^6+2x^3)$$ $$=\frac16 \frac{(x^6+2x^3)^{4/3}}{4/3}+C$$ $$=\frac{(x^6+2x^3)^{4/3}}{8}+C$$