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

1 Answer
May 22, 2016

By converting the fraction into a sum of independent factors

Explanation:

We know how to integrate functions of the form #1/x^n# or #1/(x-a)^n# so we want to transform the complicated fraction into a sum of simple fractions. We write
#I = A/(x-1) + B/(x-1)^2 + C/(x^2 + 1)#
The constant term in the numerator is #B + C-A# and is equal to #-1#
The #x^4# term is #B+C+A# and must be zero. This gives us
#-2A = -1# or #A=-1/2#
The #x^3# term is #-2C + A + B = 0#

Next #-2C + A +B +2C = 0 + 2(-A-B)# so
#B = -A = 1/2# and thus #C=0#

Thus #I = -(1/2)1/(x-1) + (1/2)1/(x-1)^2 #

Integral #= -1/2 ln(x-1) -1/2(x-1)^2#