How do you find the integral `ln(x^2+x+1)`?

`intln(x^2+x+1)dx`

apply Integration by Parts

`u=ln(x^2+x+1)`

`du=(2x+1)/(x^2+x+1)dx`

`dv=dx`

`v=x`

`intln(x^2+x+1)dx=xln(x^2+x+1)-int(x(2x+1))/(x^2+x+1)dx`
`rarr``(1)`

Let `I=int(x(2x+1))/(x^2+x+1)dx=`

simplify

`int(2x^2+x)/(x^2+x+1)dx`=`int(x^2+x+1)/(x^2+x+1)dx+int(x^2-1)/(x^2+x+1)dx`

`=x+int(x^2-1)/(x^2+x+1)dx`

let `r=int(x^2-1)/(x^2+x+1)dx`

completing the square of the denominator :
`r=int(x^2-1)/((x+1/2)^2+3/4)dx`

apply trigonometric substitution

`x+1/2=sqrt(3)/2tantheta`
`dx=sqrt(3)/2sec^2theta*d(theta)`
`x^2=3/4tan^2theta-sqrt(3)/2tantheta+1/4`

substitute

`int(x^2-1)/((x+1/2)^2+3/4)dx=int((3/4tan^2theta-sqrt(3)/2tantheta+1/4)-1)/(3/4tan^2theta+3/4)sqrt(3)/2sec^2thetad(theta)`

simplify where `tan^2theta+1=sec^2theta`

`4/3*sqrt(3)/2int(3/4tan^2theta-sqrt(3)/2tantheta-3/4)d(theta)`

simplify again using the relation `tan^2theta+1=sec^2theta`

`2/sqrt3int(3/4sec^2theta-sqrt3/2tantheta-3/2)d(theta)`=`sqrt3/2tantheta+lncostheta-sqrt3theta+C`

reverse the trigonometric substitution

`theta=tan^-1(2/sqrt(3)(x+1/2))`

so,
`r=int(x^2-1)/(x^2+x+1)dx`=`(x+1/2)+lncostan^-1(2/sqrt(3)(x+1/2))-sqrt3tan^-1(2/sqrt(3)(x+1/2))+C`

substituting in `I`:

`I=x+(x+1/2)+lncostan^-1(2/sqrt(3)(x+1/2))-sqrt3tan^-1(2/sqrt(3)(x+1/2))+C`

and substituting with the value of I in (1)

your final integration result will be:

`intln(x^2+x+1)dx`=`xln(x^2+x+1)-x-(x+1/2)-lncostan^-1(2/sqrt(3)(x+1/2))+sqrt3tan^-1(2/sqrt(3)(x+1/2))+C`