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

Answer 1

Emilia Aguilar

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#

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Answer 2

Christopher Allers

To integrate ∫(x^2)/(1-x^3) dx, use the substitution method. Let u = 1 - x^3, then du = -3x^2 dx. Solve for dx to get dx = du/(-3x^2). Substitute these into the integral:

∫(x^2)/(1-x^3) dx = ∫(-1/3)*(1/u) du = (-1/3) ln|u| + C

Now, substitute back u = 1 - x^3:

= (-1/3) ln|1 - x^3| + C

Therefore, the integral of (x^2)/(1-x^3) dx is (-1/3) ln|1 - x^3| + C.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.