$\displaystyle \int x^5(x^3+1)^{1/3}dx$

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- Feb 17th 2009, 03:37 PMLarriotoIndefinite integral
$\displaystyle \int x^5(x^3+1)^{1/3}dx$

- Feb 17th 2009, 04:23 PMScott H
Integrands with radicals can often be simplified by substituting $\displaystyle u$ for the radical. This integral can be calculated by letting $\displaystyle u=(x^3+1)^{\frac{1}{3}}$.

The next step is to cube both sides, $\displaystyle u^3=x^3+1$, and relate the differentials.