I am trying to evaluate:
$\int (x^6+x^3)\sqrt[3]{x^3+2} \ \ dx$
My solution:
$\int (x^5+x^2)\sqrt[3]{x^6+2x^3} \ \ dx$
Let $(x^6+2x^3) = t^3 \ \ \text{and} \ \ (x^5+x^2) \ \ dx = \frac{1}{2}t^2 \ \ dt$
$\frac{1}{2}\int t^2\cdot t \ \ dt = \frac{1}{2}.\frac{t^4}{4}+C $
So $\int (x^5+x^2)\sqrt[3]{x^6+2x^3} \ \ dx = \frac{1}{8}(x^6+2x^3)^{{4}/{3}}+C$
Is that right? And is there a different way ?