∫(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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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?
It's x^3 though, not x^2
$\displaystyle \frac{x^3}{2x} = \frac{1}{2}x^2$ I forgot to put the constant in my previous post.
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