<s>[INST] Evaluating $\int (x^6+x^3)\sqrt[3]{x^3+2}dx$ 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 ?
 [/INST] The $t$ stuff is not necessary. You can directly let $u=x^6+2x^3$. Then $(x^5+x^2)\,dx=\frac{1}{6}\,du$. But the initial step was the key one.
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