Let's do the partial fraction decomposition
(x+2)/((x^2+x+7)(x+1))=(Ax+B)/(x^2+x+7)+C/(x+1)x+2(x2+x+7)(x+1)=Ax+Bx2+x+7+Cx+1
=((Ax+B)(x+1)+C(x^2+x+7))/((x^2+x+7)(x+1))=(Ax+B)(x+1)+C(x2+x+7)(x2+x+7)(x+1)
So x+2=(Ax+B)(x+1)+C(x^2+x+7)x+2=(Ax+B)(x+1)+C(x2+x+7)
Let x=-1x=−1 then 1=0+7C1=0+7C =>⇒C=1/7C=17
let x=0x=0 then 2=B+7C2=B+7C 2=B+12=B+1 =>⇒ B=1B=1
Coefficients of x^2x2
0=A+C0=A+C =>⇒ A=-C=-1/7A=−C=−17
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)∫(x+2)dx(x2+x+7)(x+1)=∫((−17)x+1)dxx2+x+7+∫(17)dxx+1
=intdx/(7(x+1))-int((x-7)dx)/(7(x^2+x+7))=∫dx7(x+1)−∫(x−7)dx7(x2+x+7)
intdx/(7(x+1))=ln(x+1)/7∫dx7(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))∫(x−7)dx7(x2+x+7)=∫xdx7(x2+x+7)−∫dx(x2+x+7)
=int((2x+1)dx)/(14(x^2+x+7))-int(15dx)/(14(x^2+x+7))=∫(2x+1)dx14(x2+x+7)−∫15dx14(x2+x+7)
int((2x+1)dx)/(14(x^2+x+7))=(1/14)ln(x^2+x+7))∫(2x+1)dx14(x2+x+7)=(114)ln(x2+x+7))
Now remains int(15dx)/(14(x^2+x+7))=15/14intdx/(x^2+x+1/4+27/4)∫15dx14(x2+x+7)=1514∫dxx2+x+14+274
=15/14intdx/(((x+1/2)^2)+27/4)=1514∫dx((x+12)2)+274
=15/14intdx/((x+1/2)^2/(3sqrt3/2)^2+1=1514∫dx(x+12)2(3√32)2+1
=15/14*2/(3sqrt3)arctan((x+1/2)/(3sqrt3/2))1514⋅23√3arctan⎛⎜⎝x+123√32⎞⎟⎠
=(5/(7sqrt3))arctan((x+1/2)/(3sqrt3/2))=(57√3)arctan⎛⎜⎝x+123√32⎞⎟⎠
