Let us first find partial fractions of #1/((x-1)(x+1)^2)# and for this let
#1/((x-1)(x+1)^2)hArrA/(x-1)+B/(x+1)+C/(x+1)^2# or
#1/((x-1)(x+1)^2)hArr(A(x+1)^2+B(x-1)(x+1)+C(x-1))/((x-1)(x+1)^2)# or
#1/((x-1)(x+1)^2)hArr(A(x^2+2x+1)+B(x^2-1)+C(x-1))/((x-1)(x+1)^2)# or
#1/((x-1)(x+1)^2)hArr((A+B)x^2+(2A+C)x+(A-B-C))/((x-1)(x+1)^2)#
Therefore #A+B=0# and #2A+C=0# and #A-B-C=1#
From these #B=-A#, #C=-2A#, Hence #A-(-A)-(-2A)=1# or #4A=1#
Hence #A=1/4#, #B=-1/4# and #C=-1/2#
Hence #int1/((x-1)(x+1)^2)dx#
= #int[1/(4(x-1))-1/(4(x+1))-1/(2(x+1)^2)]dx#
= #1/4ln(x-1)-1/4ln(x+1)+1/(2(x+1))+c#
