I decomposed integrand into basic fractions,
#(x^3-x-1)/[x*(x^4-1)]#
=#(x^3-x-1)/[x(x+1)(x-1)(x^2+1)]#
=#A/x+B/(x+1)+C/(x-1)+(Dx+E)/(x^2+1)#
After expanding denominators,
#A(x+1)(x-1)(x^2+1)+Bx(x-1)(x^2+1)+Cx(x+1)(x^2+1)+(Dx+E)x(x^2-1)=x^3-x-1#
Set #x=-1#, #4B=-1 or B=-1/4#
Set #x=0#, #-A=-1 or A=1#
Set #x=1#, #4C=-1 or C=-1/4#
Set #x=i#, #(E+Di)*(-2i)=-1-2i or E+Di=1-1/2*i#. So #D=-1/2 and E=1#
Hence,
#(x^3-x-1)/[x*(x^4-1)]#
=#1/x-1/4*1/(x+1)-1/4*(x-1)-1/2*(x-2)/(x^2+1)#
Hence,
#int (x^3-x-1)/[x*(x^4-1)]*dx#
=#int dx/x#-#1/4##dx/(x+1)#-#1/4##dx/(x-1)#-#1/2#*#int ((x-2)*dx)/(x^2+1)#
=#Lnx-1/4*Ln(x+1)-1/4*Ln(x-1)#-#1/4#*#int (2x*dx)/(x^2+1)#+#dx/(x^2+1)#
=#Lnx-1/4*Ln(x+1)-1/4*Ln(x-1)-1/4*Ln(x^2+1)+arctanx+C#
=#Lnx-1/4*Ln(x^4-1)+arctanx+C#
