Let's do the partial fraction decomposition
(x+2)/((x^2+x+7)(x+1))=(Ax+B)/(x^2+x+7)+C/(x+1)
=((Ax+B)(x+1)+C(x^2+x+7))/((x^2+x+7)(x+1))
So x+2=(Ax+B)(x+1)+C(x^2+x+7)
Let x=-1 then 1=0+7C =>C=1/7
let x=0 then 2=B+7C 2=B+1 => B=1
Coefficients of x^2
0=A+C => A=-C=-1/7
so the integral becomes
int((x+2)dx)/((x^2+x+7)(x+1))=int(((-1/7)x+1)dx)/(x^2+x+7)+int((1/7)dx)/(x+1)
=intdx/(7(x+1))-int((x-7)dx)/(7(x^2+x+7))
intdx/(7(x+1))=ln(x+1)/7
int((x-7)dx)/(7(x^2+x+7))=int(xdx)/(7(x^2+x+7))-int(dx)/((x^2+x+7))
=int((2x+1)dx)/(14(x^2+x+7))-int(15dx)/(14(x^2+x+7))
int((2x+1)dx)/(14(x^2+x+7))=(1/14)ln(x^2+x+7))
Now remains int(15dx)/(14(x^2+x+7))=15/14intdx/(x^2+x+1/4+27/4)
=15/14intdx/(((x+1/2)^2)+27/4)
=15/14intdx/((x+1/2)^2/(3sqrt3/2)^2+1
=15/14*2/(3sqrt3)arctan((x+1/2)/(3sqrt3/2))
=(5/(7sqrt3))arctan((x+1/2)/(3sqrt3/2))
