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

1 Answer
Roy E.
Dec 7, 2016

#ln(x-1)-2/(x-1)-1/(2(x-1)^2)+C#

Explanation:

Assume that there exist #A, B and C# such that:
#x^2/(x-1)^3-=A/((x-1))+B/(x-1)^2+C/(x-1)^3#.
To find #A#, #B# and #C# substitute any three different numbers other than #-1# and solve the resulting set of three linear equations. For example, set #x=0#, #2# and #-1#:
#0=A/-1+B+C/-1#
#4=A+B+C#
#-1/8=A/-2+B/4-C/8#
giving #A=1#, #B=2# and #C=1#.
So the required integral is:
#int1/(x-1)+2/(x-1)^2+1/(x-1)^3dx#
which gives the answer above.

Alternatively, instead of using partial fractions, substitute #u=x-1# to get #int1/u+2/u^2+1/u^3du#.

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