Improper integral of natural log over a quadratic I need to evaluate 
$$\int\limits_0^{+\infty}\frac{\ln{x}}{x^2+x+1}\,\mathrm{d}x\,.$$
I don't know how to integrate this, and for the most part, I don't even think it is expressible as elementary functions. In that case, how would I even manipulate the integral using some $u$-substitution to transform this into some integrable function? Or can this all be done without actual integration, and just some clever substitution to somehow find a multiple of this integral's value?
 A: First of all, the integral is convergent because :


*

*$\displaystyle{\frac{\ln(x)}{x^2+x+1}\underset{0}{\sim}\ln(x)}$

*$\displaystyle{\frac{\ln(x)}{x^2+x+1}\underset{+\infty}{=}o\left(\frac1{x^{3/2}}\right)}$
Now consider change of variable $\displaystyle{t=\frac 1x}$
You will get  :
$$\int_1^\infty\frac{\ln(x)}{x^2+x+1}\,dx=\int_1^0\frac{-\ln(t)}{\frac1{t^2}+\frac 1t+1}\frac{-dt}{t^2}=-\int_0^1\frac{\ln(t)}{t^2+t+1}\,dt$$
This proves that :
$$\boxed{\int_0^{+\infty}\frac{\ln(x)}{x^2+x+1}\,dx=0}$$

It should be added that, more generally (and for the same reasons) :
$$\forall a\in(-2,+\infty),\,\int_0^{+\infty}\frac{\ln(x)}{x^2+ax+1}\,dx=0$$
A: Substitute $u=\log(x),$ giving $$ \int_0^\infty \frac{\log(x)}{x^2+x+1}dx = \int_{-\infty}^\infty \frac{u}{e^u+1+e^{-u}}du = 0$$ since the integrand is odd. (And the integral exists since the integrand decays exponentially in both directions.)
A: Considering $$\int_{0}^{\infty}\frac{\ln{x}}{x^2+x+1}\,dx=\int_{0}^{1}\frac{\ln{x}}{x^2+x+1}\,dx+\int_{1}^{\infty}\frac{\ln{x}}{x^2+x+1}\,dx$$ For the second integral, change variable $x=\frac 1y$, simplify and admire !
