(7x^2+8)/((x-1)(x^2+x+1))=A/(x-1)+(Bx+C)/(x^2+x+1).
7x^2+8=Ax^2+Ax+A+(Bx+C)(x-1)=Ax^2+Ax+A+Bx^2+(C-B)x-C.
Equating coefficients:
x^2: A+B=7, B=7-A
x: A+C-B=0
Constant: A-C=8, C=A-8
A+A-8-7+A=0, 3A=15, A=5; B=7-5=2; C=5-8=-3.
Therefore we have to integrate: 5/(x-1)+(2x-3)/(x^2+x+1).
The first term integrated is 5ln|x-1|.
2x-3=2x+1-4; the differential of x^2+x+1 is 2x+1, so ∫(2x-3)dx/(x^2+x+1)=ln|x^2+x+1|-4∫dx/(x^2+x+1).
x^2+x+1=x^2+x+1/4 -1/4+1=(x+1/2)^2+3/4.
The integral of 1/(x^2+a^2)=(1/a)arctan(x/a). If a=√3/2 and x is replaced by x+1/2, the integral is (2/√3)arctan((2x+1)/√3). We can rationalise 1/√3 as √3/3. Therefore this integral is (2√3/3)arctan((2x+1)√3/3).
The final integral is 5ln|x-1|+ln|x^2+x+1|-(8√3/3)arctan((2x+1)√3/3)+C, where C is constant of integration.