How do you integrate #int (x^2)/(1-x^3) dx#?

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Answer 1 · 2015-06-08T01:00:12

You can do a u-substitution.

Notice how this can be written as:

#int x^2*(1/(1-x^3))dx#

#d/(dx)[-x^3] = -3x^2dx#, so #d/(dx)[(-1/3)*(-x^3)] = x^2dx#.

Let:
#u = 1-x^3#
#du = -3x^2dx#

#=> -1/3int (1/(1-x^3))*(-3x^2)dx#

#= -1/3int1/udu#

#= -1/3ln|u| + C#

#= -1/3ln|1-x^3| + C#