Question

How do you integrate `int x^2 / (x-1)^3` using partial fractions?

Answer 1

I got `ln|x-1| - 2/(x-1) - 1/(2(x-1)^2) + C`.

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With partial fractions where the denominator has a multiplicity of `3` (that is, it is a perfect cube), this factors into:

`int (x^2)/(x-1)^3dx = int A/(x-1) + B/(x-1)^2 + C/(x-1)^3dx`

Now, we should get common denominators so that we can set the resultant fraction equal to the starting integrand.

That way, we can equate each coefficient to the numerator coefficients and find `A`, `B`, and `C`.

Ignoring the integral symbols for now, we can focus on the integrand:

`= (A(x-1)^2)/(x-1)^3 + (B(x-1))/(x-1)^3 + C/(x-1)^3`

Combine the fractions:

`= (A(x-1)^2 + B(x-1) + C)/(x-1)^3`

Distribute the numerator terms:

`= (Ax^2 - 2Ax + A + Bx - B + C)/cancel((x-1)^3) = (x^2)/cancel((x-1)^3)`

Next, we can rearrange the terms to turn this into the form `a_0x^2 color(highlight)(+) a_1x color(highlight)(+) a_2`, which is the standard form of a quadratic equation.

Remember that this form must include the `color(highlight)("addition")` of all ordered terms as a group; no subtractions!

`(A)x^2 \mathbf(color(highlight)(+)) (-2A + B)x \mathbf(color(highlight)(+)) (A - B + C) = x^2`

Thus, we have the following system of equations:

`color(green)(A = 1)`
`-2A + B = 0`
`A - B + C = 0`

Knowing `A` already, we can easily solve this to get:

`2A = B => color(green)(B = 2)`
`A - B + C = 0 => 1 - 2 + C = 0 => color(green)(C = 1)`

Thus, our resultant integrals are calculated as follows:

`color(blue)(int (x^2)/(x-1)^3dx) = int 1/(x-1) + 2/(x-1)^2 + 1/(x-1)^3dx`

`= int 1/(x-1)dx + 2int1/(x-1)^2dx + int1/(x-1)^3dx`

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

`= ln|x-1| + (2(x-1)^(-1))/(-1) + ((x-1)^(-2))/(-2)`

`= color(blue)(ln|x-1| - 2/(x-1) - 1/(2(x-1)^2) + C)`