I decomposed integrand into basic fractions,
(x+1)/[(x^2-2)*(3x+1)]=(Ax+B)/(x^2-2)+C/(3x-1)x+1(x2−2)⋅(3x+1)=Ax+Bx2−2+C3x−1
After expanding denominator,
(Ax+B)(3x-1)+C(x^2-2)=x+1(Ax+B)(3x−1)+C(x2−2)=x+1
Set x=1/3x=13, -17C/9=4/3−17C9=43, so C=-12/17C=−1217
Set x=0x=0, -B-2C=1−B−2C=1, so B=-2C-1=7/17B=−2C−1=717
Set x=1x=1, 2A+2B-C=22A+2B−C=2, so A=1/2*(2+C-2B)=4/17A=12⋅(2+C−2B)=417
Hence,
int (x+1)/[(x^2-2)*(3x+1)]*dx∫x+1(x2−2)⋅(3x+1)⋅dx
=1/17*int ((4x+7)*dx)/(x^2-2)-1/17*int (12*dx)/(3x-1)117⋅∫(4x+7)⋅dxx2−2−117⋅∫12⋅dx3x−1
=2/17*int (2x*dx)/(x^2-2)+(7sqrt(2))/68*int (2sqrt2*dx)/(x^2-2)-4/17*int (3*dx)/(3x-1)217⋅∫2x⋅dxx2−2+7√268⋅∫2√2⋅dxx2−2−417⋅∫3⋅dx3x−1
=2/17*ln(x^2-2)+(7sqrt(2))/68*Ln((x-sqrt2)/(x+sqrt2))-4/17*Ln(3x-1)+C217⋅ln(x2−2)+7√268⋅ln(x−√2x+√2)−417⋅ln(3x−1)+C
