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!