Let's rewrite the denominator as y^3-1=(y-1)(y^2+y+1)
And the decompsition into partial fractions
(y^2+1)/(y^3-1)=A/(y-1)+(Bx+C)/(y^2+y+1)
=(A(y^2+y+1)+(By+C)(y-1))/((y-1)(y^2+y+1))
So, y^2+1=A(y^2+y+1)+(By+C)(y-1)
Let y=0 then, 1=A-C
Coefficients of y^2, 1=A+B
Coefficients of y, 0=A-B+C
Solving for A, B and C, we get
A=2/3, B=1/3 and C=-1/3
Then
int((y^2+1)dy)/(y^3-1)=int(2dy)/(3(y-1))+int((1/3y-1/3)dy)/(y^2+y+1
=int(2dy)/(3(y-1))+int((2y+1)dy)/(6(y^2+y+1))-int(dy)/(2(y^2+y+1))
We have to find 3 integrals
int(2dy)/(3(y-1))=2/3ln(y-1)
int((2y+1)dy)/(6(y^2+y+1))=ln(y^2+y+1)/6
And the last integral
int(dy)/(2(y^2+y+1))=1/2intdy/(y^2+y+1/4+3/4)
=1/2intdy/((y+1/2)^2+3/4)
=1/2*4/3intdy/(((y+1/2)/(sqrt3/2))^2+1)
Let u=(y+1/2)/(sqrt3/2)=(2y+1)/sqrt3
then du=2dy/sqrt3 =>dy=(sqrt3du)/2
so =2/3int(sqrt3/2du)/(u^2+1)=1/sqrt3int(du)/(u^2+1)
=1/sqrt3arctanu=1/sqrt3arctan((2y+1)/sqrt3)
So putting it all together
int((y^2+1)dy)/(y^3-1)=2/3ln(y-1)+ln(y^2+y+1)/6-1/sqrt3arctan((2y+1)/sqrt3)
