Perform the decomposition into partial fractions
#1/(x^2-1)^2=1/((x+1)^2(x-1)^2)=A/(x+1)^2+B/(x+1)+C/(x-1)^2+D/(x-1)#
#=(A(x-1)^2+B(x+1)(x-1)^2+C(x+1)^C2+D(x-1)(x+1)^2)/(((x+1)^2(x-1)^2))#
The denominators are the same, compare the numerators
#1=A(x-1)^2+B(x+1)(x-1)^2+C(x+1)^2+D(x-1)(x+1)^2#
Let #x=-1#, #=>#, #1=4A#, #=>#, #A=1/4#
Let #x=1#, #=>#, #1=4C#, #=>#, #C=1/4#
Coefficients of #x^2#
#0=A-B+C+D#, #=>#, #0=1/4-B+1/4+D#, #=>#, #B-D=1/2#
Coefficients of #x^3#
#0=B+D#, #=>#, #D=-B#, #=>#
#B=1/4#,
#D=-1/4#
Therefore,
#1/(x^2-1)^2=(1/4)/(x+1)^2+(1/4)/(x+1)+(1/4)/(x-1)^2+(-1/4)/(x-1)#
so,
#int(dx)/(x^2-1)^2=int(1/4dx)/(x+1)^2+int(1/4dx)/(x+1)+int(1/4dx)/(x-1)^2+int(-1/4dx)/(x-1)#
#=-1/(4(x+1))+1/4ln(|x+1|)-1/(4(x-1))-1/4ln(|x-1|)+C#
