Question

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

Answer 1

`1/4 ln|x-1| + 3/4 ln|x+1| + 1/(2(x+1)) + C`

Explanation:

Let `x^2/((x-1)(x^2+2x+1))-=(1/4)/(x-1)+(Bx+C)/(x+1)^2` for all `x`.
(The `1/4` comes from the cover-up rule: `1^2/(1^2+2xx1+1)`.)
Then, multiplying out,
`4x^2-=x^2+2x+1 + 4Bx^2+4(C-B)x-4C`
Collecting powers of `x`:
`4x^2-=(1+4B)x^2+(2+4C+4B)x+(1-4C)`
Equating powers of `x^2` on both sides of the identity:
`4=1+4B` hence `B=3/4`.
Equating constants on both sides gives `C=1/4`.
Hence the integral is:
`(1/4)int(1/(x-1)+(3x+1)/(x+1)^2)dx`
`=1/4int(1/(x-1)+(3x+3-2)/(x+1)^2)dx`
`=1/4int(1/(x-1)+(3x+3)/(x+1)^2-2/(x+1)^2)dx`
`=1/4ln|x-1|+3/8 ln(x+1)^2+1/(2(x+1))+C`