We know that,
#color(blue)((1) int(f^'(x))/(f(x))dx=ln|f(x)|+c#
#color(red)((2) int1/(X^2+A^2)dX=1/A tan^-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-3/2int1/(x^2+x+2)dx#
#=x-1/2intcolor(blue)((d/(dx)((x^2+x+2)))/(x^2+x+2)dx)-
3/2int1/(x^2+x+1/4+7/4)dx#
Using #(1)# we get
#=x-1/2color(blue)(ln|(x^2+x+2)|)-3/2intcolor(red)(1/((x+1/2)^2+(sqrt7/2)^2)dx#
Using #(2)# we get
#=x-1/2ln|(x^2+x+2)|-3/2*color(red)(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#