∫(x^3)(x^2-1)^(3/2)dx

I tried u=x^2 - 1, but then I get (1/2)du=xdx, and I have an x^3, so that's not working...what should I do?

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- Dec 1st 2008, 06:14 PMSusaludaUsing u-substitution, integrate ∫(x^3)(x^2-1)^(3/2)dx
∫(x^3)(x^2-1)^(3/2)dx

I tried u=x^2 - 1, but then I get (1/2)du=xdx, and I have an x^3, so that's not working...what should I do? - Dec 1st 2008, 06:20 PMChop Suey
On the contrary, $\displaystyle u = x^2 - 1$ is an ideal substitution.

$\displaystyle du = 2x~dx$ yields

$\displaystyle \int (x^2)(x^2-1)^{\frac{3}{2}}~du = \int (u+1)u^{\frac{3}{2}}~du$

Do you see why? - Dec 1st 2008, 06:24 PMSusaluda
It's x^3 though, not x^2

- Dec 1st 2008, 06:27 PMChop Suey
$\displaystyle \frac{x^3}{2x} = \frac{1}{2}x^2$

I forgot to put the constant in my previous post.