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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 ?

  • 1
    Looks right to me. You can verify it by differentiating your answer.2012-07-17

1 Answers 1

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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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    As is often the case, I didn't read, it is usually easier to solve.2012-07-17