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How do you integrate #x^2/(x^3+1)#?

↳Redirected from "How do I change #int_0^1int_0^sqrt(1-x^2)int_sqrt(x^2+y^2)^sqrt(2-x^2-y^2)xydzdydx# to cylindrical or spherical coordinates?"

1 Answer

Mar 16, 2015

#int x^2/(x^3+1) dx = 1/3int(3x^2)/(x^3+1)dx#

Now you have #(f'(x))/f(x)# form.

Integral is : #1/3ln(|\x^3+1|)+C#

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