I decomposed integrand into basic fractions,
(x^3-x-1)/[x*(x^4-1)]x3−x−1x⋅(x4−1)
=(x^3-x-1)/[x(x+1)(x-1)(x^2+1)]x3−x−1x(x+1)(x−1)(x2+1)
=A/x+B/(x+1)+C/(x-1)+(Dx+E)/(x^2+1)Ax+Bx+1+Cx−1+Dx+Ex2+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-1A(x+1)(x−1)(x2+1)+Bx(x−1)(x2+1)+Cx(x+1)(x2+1)+(Dx+E)x(x2−1)=x3−x−1
Set x=-1x=−1, 4B=-1 or B=-1/44B=−1orB=−14
Set x=0x=0, -A=-1 or A=1−A=−1orA=1
Set x=1x=1, 4C=-1 or C=-1/44C=−1orC=−14
Set x=ix=i, (E+Di)*(-2i)=-1-2i or E+Di=1-1/2*i(E+Di)⋅(−2i)=−1−2iorE+Di=1−12⋅i. So D=-1/2 and E=1D=−12andE=1
Hence,
(x^3-x-1)/[x*(x^4-1)]x3−x−1x⋅(x4−1)
=1/x-1/4*1/(x+1)-1/4*(x-1)-1/2*(x-2)/(x^2+1)1x−14⋅1x+1−14⋅(x−1)−12⋅x−2x2+1
Hence,
int (x^3-x-1)/[x*(x^4-1)]*dx∫x3−x−1x⋅(x4−1)⋅dx
=int dx/x∫dxx-1/414dx/(x+1)dxx+1-1/414dx/(x-1)dxx−1-1/212*int ((x-2)*dx)/(x^2+1)∫(x−2)⋅dxx2+1
=Lnx-1/4*Ln(x+1)-1/4*Ln(x-1)lnx−14⋅ln(x+1)−14⋅ln(x−1)-1/414*int (2x*dx)/(x^2+1)∫2x⋅dxx2+1+dx/(x^2+1)dxx2+1
=Lnx-1/4*Ln(x+1)-1/4*Ln(x-1)-1/4*Ln(x^2+1)+arctanx+Clnx−14⋅ln(x+1)−14⋅ln(x−1)−14⋅ln(x2+1)+arctanx+C
=Lnx-1/4*Ln(x^4-1)+arctanx+Clnx−14⋅ln(x4−1)+arctanx+C
