We know that,
color(red)((1)int(f'(x))/(f(x))dx=ln|f(x)|+c
color(blue)((2)int1/(X^2+A^2)dX=1/Atan^-1(X/A)+c
Here,
I=intx^2/(x^2+x+2)dx
=int((x^2+x+2)-(x+2))/(x^2+x+2)dx
=int(x^2+x+2)/(x^2+x+2)dx-int(x+2)/(x^2+x+2)dx
=int1dx-1/2int(2x+4)/(x^2+x+2)dx
=x-1/2int(2x+1+3)/(x^2+x+2)dx
=x-1/2int(2x+1)/(x^2+x+2)dx-1/2int3/(x^2+x+2)dx
=x-1/2color(red)(int(d/(dx)
(x^2+x+2))/(x^2+x+2)dx)-3/2int1/(x^2+x+1/4+7/4)dx
=x-1/2color(red)(ln|x^2+x+2|)-3/2color(blue)(int1/((x+1/2)^2+
(sqrt7/2)^2)dx
=x-1/2ln|x^2+x+2|-3/2*color(blue)
(1/(sqrt7/2)tan^-1((x+1/2)/(sqrt7/2))+C
=x-1/2ln|x^2+x+2|-3/sqrt7tan^-1((2x+1)/sqrt7)+C