How do you integrate `int (x-1)/( x^4 (x-1)^2)` using partial fractions?

1) I used basic fractions method for integration.

`int ((x-1) dx)/[x^4*(x-1)^2]`

`=int dx/[x^4*(x-1)]`

I decomposed integrand into basic fractions,

`1/[x^4*(x-1)]=A/(x-1)+B/x+C/x^2+D/x^3+E/x^4`

`A*x^4+B*x^3*(x-1)+C*x^2*(x-1)+D*x*(x-1)+E*(x-1)=1`

Set `x=0`, `-E=-1` or `E=-1`

Set `x=1`, `A=1`

I took differentiaton both sides,

`4A*x^3+B*(4x^3-3x^2)+C*(3x^2-2x)+D*(2x-1)+E=0`

Set `x=0`, `-D+E=0` or `D=E=-1`

I took differentiaton both sides,

`12A*x^2+B*(12x^2-6x)+C*(6x-2)+2D=0`

Set `x=0`, `-2C+2D=0` or `C=D=-1`

I took differentiaton both sides,

`24A*x+B*(24x-6)+6C=0`

Set `x=0`, `-6B+6C=0` or `B=C=-1`

Thus,

`int dx/[x^4*(x-1)]`

=`int dx/(x-1)`-`int dx/x`-`int dx/x^2`-`int dx/x^3`-`int dx/x^4`

=`Ln(x-1)-Lnx+x^(-1)+1/2*x^(-2)+1/3*x^(-3)+C`

`int 1/(x^4(x-1))dx`
`= -int 1/(x^4(1-x))dx`
`=-int(1/x^4)(1+x+x^2+x^3+x^4...)dx` (for suitable `x`, sum to infinity of GP with common ratio `x`, backwards, or binomial expansion with `n=-1`)
`=-intx^-4+x^-3+x^-2+x^-1+1+ x+x^2+x^3+...dx`
`=1/3x^3+1/2x^-2+x^-1-ln|x|-int1/(1-x) dx` (sum to infinity of GP again, forwards)
`=1/3x^3+1/2x^-2+x^-1-ln|x|+ln|x-1|+c`

The binomial expansion above works only for `|x|<1`. To deal with `|x|>1`:
`int 1/(x^4(x-1))dx`
`=int 1/(x^5(1-1/x))dx`
Using the Binomial Theorem, or sum to infinity of GP with first term 1 and common ratio `x^-1`:
`=intx^-5(1+x^-1+x^-2+x^-3...)dx` provided `|x|>1`
`=int x^-5+x^-6+x^-7... dx`
`=int x^-1+x^-2+x^-3+x^-4+x^-5... dx -`
`\ \ \ \ \ int x^-1+x^-2+x^-3+x^-4dx`

Using sum to infinity of GP or binomial theorem:
`=intx^-1/(1-x^-1)dx - int x^-1+x^-2+x^-3+x^-4dx`
`=int 1/(x-1)dx - int x^-1+x^-2+x^-3+x^-4dx`
`=ln|x-1|-ln|x| + 1/x+1/2x^2+1/3x^3+c`
which is the same as the result for `|x|<1`. Good!