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)#
