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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 ?

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Just a note: usually $\mathbb{C}$ denotes the complex numbers, while most use just C or K to indicate a constant for integrals. – Joe Jul 17 '12 at 4:06
Looks right to me. You can verify it by differentiating your answer. – Shahab Jul 17 '12 at 4:26
up vote 4 down vote accepted

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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Nice trick ,but i did it in the solution – Frank Jul 17 '12 at 4:08
As is often the case, I didn't read, it is usually easier to solve. – André Nicolas Jul 17 '12 at 4:11

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