Hi, I'm new (and sorry for my bad english ), I'm trying to solve this complex integral $\displaystyle \int_{0}^{+\infty} \frac{log x}{(2x+1)(x^{2}+x+1)} dx$ and the result should be $\displaystyle \frac{1}{27}(-\pi^2-9log^2(2))$. I write what I did:

The function is integrable and it is not even, and I choose the auxiliary function $\displaystyle f(z)=\frac{log^{2} z}{(2z+1)(z^2+z+1)}$ with the square natural log.

The regular path I choose is the superior semicircle excluding singularity points: $\displaystyle 0$ and $\displaystyle -1/2$. The two poles are $\displaystyle -1/2+i\sqrt{3}/2$ (which falls inside) and $\displaystyle -1/2-i\sqrt{3}/2$.

$\displaystyle 2{\pi}{i}Res(f, -\frac{1}{2}+i\frac{\sqrt{3}}{2})=\int_{+\partial{T }}f(z)dz=\int_{-R}^{-\delta -\frac{1}{2}}\frac{{(log|z|+iarg(z))}^{2}}{(2z+1)(z ^2+z+1)}dz - \int_{+\Gamma\delta}f(z)dz +\int_{-\frac{1}{2}+\delta}^{\epsilon}\frac{{(log|z|+iarg( z))}^{2}}{(2z+1)(z^2+z+1)}dz - \int_{+\Gamma\epsilon}f(z)dz + \int_{+\epsilon}^{+R}\frac{{(log|z|+iarg(z))}^{2}} {(2z+1)(z^2+z+1)}dz + \int_{+\Gamma{R}}f(z)dz$.

The integral calculated on the semicicrcle centred in $\displaystyle 0$ is $\displaystyle 0$ and also the integral on the big semicircle is $\displaystyle 0$ while the integral on the semicircle centred in $\displaystyle -1/2$ is $\displaystyle \frac{2}{3}(i\pi-log(2))^{2}$. Finally the residue on the pole is $\displaystyle Res(f, \frac{-1}{2}+i\frac{\sqrt{3}}{2})=\frac{4}{27}{\pi}^{2}$ so:

$\displaystyle 2{\pi}{i}\frac{4}{27}{\pi}^{2}=\frac{2}{3}(i\pi-log(2))^{2} + \int_{-\infty}^{-\frac{1}{2}}\frac{{(log|x|+i{\pi})}^{2}}{(2x+1)(x^ 2+x+1)}dx+\int_{-\frac{1}{2}}^{0}\frac{{(log|x|+i{\pi})}^{2}}{(2x+1 )(x^2+x+1)}dx$$$\displaystyle $ + \int_{0}^{+\infty}\frac{log^{2}x}{(2x+1)(x^2+x+1)} dz$

$\displaystyle 2{\pi}{i}\frac{4}{27}{\pi}^{2}= \frac{2}{3}(i\pi-log(2))^{2} + \int_{-\infty}^{0}\frac{{(log(-x)+i{\pi}))}^{2}}{(2x+1)(x^2+x+1)}dx+ \int_{0}^{+\infty}\frac{log^{2}x}{(2x+1)(x^2+x+1)} dz$

$\displaystyle 2{\pi}{i}\frac{4}{27}{\pi}^{2}= -\frac{2}{3}\pi^{2}+\frac{2}{3}log^{2}(2)-\frac{4}{3}i\pi log(2) + \int_{-\infty}^{0}\frac{log^{2}(-x)}{(2x+1)(x^2+x+1)}dx- \int_{-\infty}^{0}\frac{\pi^{2}}{(2x+1)(x^2+x+1)}dx+\int_ {-\infty}^{0}\frac{2log(-x)i\pi}{(2x+1)(x^2+x+1)}dx+\int_{0}^{+\infty}\frac {log^{2}x}{(2x+1)(x^2+x+1)}dx$

so:

$\displaystyle 2{\pi}{i}\frac{4}{27}{\pi}^{2}= -\frac{2}{3}\pi^{2}+\frac{2}{3}log^{2}(2)-\frac{4}{3}i\pi log(2)- \int_{-\infty}^{0}\frac{\pi^{2}}{(2x+1)(x^2+x+1)}dx+\int_ {-\infty}^{0}\frac{2log(-x)i\pi}{(2x+1)(x^2+x+1)}dx$

I change the sign of the integral I want find:

$\displaystyle 2{\pi}{i}\frac{4}{27}{\pi}^{2}= -\frac{2}{3}\pi^{2}+\frac{2}{3}log^{2}(2)-\frac{4}{3}i\pi log(2)- \int_{-\infty}^{0}\frac{\pi^{2}}{(2x+1)(x^2+x+1)}dx-\int_{0}^{+\infty}\frac{2log(x)i\pi}{(2x+1)(x^2+x+ 1)}dx$

Considering only the coefficients with $\displaystyle $-i$$.

$\displaystyle -\frac{4}{27}{\pi}^{2}-\frac{2}{3}log(2)=\int_{0}^{+\infty}\frac{log(x)}{ (2x+1)(x^2+x+1)}dx$

but this is not the result... :|

Some ideas?? Thanks for understanding