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))#
