#int (3x+6)/(x^4-2x^2+1)*dx#
=#int (3x+6)/(x^2-1)^2*dx#
=#int (3x+6)/[(x+1)*(x-1)]^2*dx#
Now I decomposed integrand into basic fractions,
#(3x+6)/[(x+1)*(x-1)]^2=A/(x+1)+B/(x+1)^2+C/(x-1)+D/(x-1)^2#
After expanding denominator,
#A(x+1)(x-1)^2+B(x-1)^2+C(x-1)(x+1)^2+D(x+1)^2=3x+6# or
#A*(x^3-x^2-x+1)+B*(x^2-2x+1)+C*(x^3+x^2-x-1)+D*(x^2+2x+1)=3x+6#
#(A+C)*x^3+(-A+B+C+D)*x^2+(-A-2B-C+2D)*x+A+B-C+D=3x+6#
After equating coefficients, I found
#A+C=0#
#-A+B+C+D=0#
#-A-2B-C+2D=3# and
#A+B-C+D=6#
After solving them, I found #A=3/2#, #C=-3/2#, #D=9/4# and #B=3/4#
Thus,
#int (3x+6)/[(x+1)*(x-1)]^2*dx#
=#3/2int (dx)/(x+1)+3/4int (dx)/(x+1)^2-3/2int (dx)/(x-1)+9/4int (dx)/(x-1)^2#
=#3/2Ln(x+1)-3/(4x+4)-3/2Ln(x-1)-9/(4x-4)+C#
=#3/2Ln((x+1)/(x-1))-(6x+3)/(2x^2-2)+C#
