#int 5/(x^2-1)^2*dx#
=#int 5/[(x+1)*(x-1)]^2*dx#
I decomposed integrand into basic fractions,
#5/[(x+1)*(x-1)]^2=A/(x+1)+B/(x+1)^2+C/(x-1)+D/(x-1)^2#
After expanding denominator,
#A*(x^3-x^2-x+1)+B*(x^2-2x+1)+C*(x^3+x^2-x-1)+D*(x^2+2x+1)=5#
Set #x=-1#, so #4B=5# or #B=5/4#
Set #x=1#, so #4D=5# or #D=5/4#
Due to denominator have double roots of #-1# and #1#, I took derivative both sides,
#A*(3x^2-2x-1)+B*(2x-2)+C*(3x^2+2x-1)+D*(2x+2)=0#
Set #x=-1#, so #4A-4B=0# or #A=B=5/4#
Set #x=1#, so #4C+4D=0# or #C=-D=-5/4#
Hence,
#int 5/[(x+1)*(x-1)]^2*dx=5/4*int (dx)/(x+1)+5/4*int (dx)/(x+1)^2-5/4*int (dx)/(x-1)+5/4*int (dx)/(x-1)^2#
=#5/4*Ln(x+1)-5/4*(x+1)^(-1)-5/4*Ln(x-1)-5/4*(x-1)^(-1)+C#
=#5/4*Ln((x+1)/(x-1))-5/2*x/(x^2-1)+C#
