Definite integral, quotient of logarithm and polynomial: $I(\lambda)=\int_0^{\infty}\frac{\ln ^2x}{x^2+\lambda x+\lambda ^2}\text{d}x$ I was thinking this integral : $$I(\lambda)=\int_0^{\infty}\frac{\ln ^2x}{x^2+\lambda x+\lambda ^2}\text{d}x$$
What I do is use a Reciprocal subsitution, easy to show that:
$$I(\lambda)=\int_0^{\infty}\frac{\ln ^2x}{\lambda ^2x^2+\lambda x+1}\text{d}x
=\frac{1}{\lambda}\int_0^{\infty}\frac{(\ln x-\ln \lambda)^2}{x^2+x+1}\text{d}x \\
=\frac{\ln ^2\lambda}{\lambda}\int_0^{\infty}\frac{1}{x^2+x+1}\text{d}x+\frac{1}{\lambda}\int_0^{\infty}\frac{\ln ^2x}{x^2+x+1}\text{d}x \\
=\frac{\ln ^2\lambda}{\lambda}\frac{2\pi}{3\sqrt{3}}+\frac{2}{\lambda}\int_0^1\frac{\ln ^2x}{x^2+x+1}\text{d}x$$
But I hav no idea on process the remaining integral:(
Anyone knows how to solve it?
THX guys! I came with another method might work for this:
Recall the handy GF $$\frac{1}{x^2+x+1}=\frac{2}{\sqrt{3}}\sum_{k=0}^{\infty}\sin \left(\frac{2\pi}{3}\left(k+1\right)\right)x^k$$
Then we have : 
$$\int_0^1\frac{\ln ^2x}{x^2+x+1}\text{d}x=\frac{2}{\sqrt{3}}\sum_{k=0}^{\infty}\frac{2}{\left(k+1\right)^3}\sin \left(\frac{2\pi}{3}\left(k+1\right)\right)x^k$$
With $$\sum_{k=1}^{\infty}\frac{\sin (kx)}{k^3}=\frac{\pi ^2}{6}x-\frac{\pi}{4}x^2+\frac{x^3}{12}$$
What can we arrived? 
 A: First, let me say that calculating this integral is not so easy, and according to my notes it evaluates to $\frac{16\pi^{3}}{81\sqrt{3}}.$
Hint: To evaluate $$\int_0^\infty \frac{(\log x)^2}{x^2+x+1}dx,$$ we take advantage of the fact that $$(a+1)^3-(a-1)^3 = 6a^2+2$$ and let $$f(x)=\frac{\left(\log x\right)^{3}}{x^{2}-x+1},$$ with a negative sign in the denominator.  Consider the integral $\oint_{\mathcal{C}}f(x)dx$ where $\mathcal{C}$ is the counter clockwise oriented keyhole contour which is a circle of radius $R$, the circle of radius $\epsilon,$ and the two branches which extend to negative infinity on the upper and lower half plane.  It is easy to see that as $R\rightarrow\infty,$ and as $\epsilon\rightarrow0,$ the two circles contribution goes to zero. In the limit, the integral on the branches becomes $$\int_{0}^{\infty}\frac{\left(\log x+\pi i\right)^{3}}{x^{2}+x+1}dx-\int_{0}^{\infty}\frac{\left(\log x-\pi i\right)^{3}}{x^{2}+x+1}dx.$$
I think you can solve it from here.
A: Here is a closed form solution for the integral
$$\int_0^1\frac{\ln ^2x}{x^2+x+1}\text{d}x=\frac{8\sqrt{3}}{243}\pi^3 \sim 1.768047624. $$
I have introduced a general technique, which is based on partial fraction combined with the use of dilogarithm function, to solve the integral

$$ \int _{a }^{b }\!{\frac {\ln  \left( tx + u \right) }{m{x}^{2}+nx+p}}{dx}.$$

In your case, instead of using the dilogarithm function, we will use the polylogarithm function $\operatorname{Li}_{s}(z)$ and the whole problem boils down to evaluate integrals of the form

$$ \int_{0}^{1} \frac{\ln(x)^2}{x-\alpha}= -2 Li_{3}\left(\frac{1}{a}\right), $$

where 

$$ \operatorname{Li}_{s}(z) = \sum_{k=1}^{\infty}\frac{z^k}{k^s}. $$ 

A: $$\int_0^1 \dfrac{\ln^2(x)}{1+x+x^2} = \int_0^1 \dfrac{1-x}{1-x^3} \ln^2(x) dx = \int_0^1 (1-x)\ln^2(x) \left(\sum_{k=0}^{\infty} x^{3k}\right) dx$$
Now note that
$$\underbrace{\int_0^1 x^n \ln^2(x)dx = \int_{-\infty}^0 e^{nt} t^2 e^t dt}_{x \mapsto e^t} = \int_0^{\infty} t^2 e^{-(n+1)t}dt = \dfrac2{(1+n)^3}$$
Hence,$$\int_0^1 \dfrac{\ln^2(x)}{1+x+x^2} dx = \sum_{k=0}^{\infty} \left(\dfrac2{(1+3k)^3} - \dfrac2{(2+3k)^3} \right) = \dfrac{8 \pi^3}{81 \sqrt3}$$
Let us call
$$\sum_{k=0}^{\infty} \dfrac1{(1+3k)^3} =f$$
and
$$\sum_{k=0}^{\infty} \dfrac1{(2+3k)^3} =g$$
We are interested in $f-g$.
We have
$$\text{Li}_3(\omega) = \sum_{k=1}^{\infty} \dfrac{\omega^k}{k^3} = \omega f + \omega^2 g + \dfrac{\zeta(3)}{27}$$
$$\text{Li}_3(\omega^2) = \sum_{k=1}^{\infty} \dfrac{\omega^{2k}}{k^3} = \omega^2 f + \omega g + \dfrac{\zeta(3)}{27}$$
where $\text{Li}_s(x)$ is the polylogarithm function defined as
$$\text{Li}_s(x) = \sum_{k=0}^{\infty} \dfrac{x^k}{k^s}$$
Polylgarithm function satisfies a nice identity namely
$$\text{Li}_n(e^{2 \pi ix}) + (-1)^n \text{Li}_n(e^{-2 \pi ix}) = - \dfrac{(2\pi i)^n}{n!}B_n(x)$$ where $B_n(x)$ are Bernoulli polynomials. Take $n=3$ and $x = 1/3$ to get that
$$\text{Li}_3(\omega) - \text{Li}_3(\omega^2) = - \dfrac{(2\pi i)^3}{3!}B_3(1/3) = - \dfrac{(2\pi i)^3}{3!} \dfrac1{27} = \dfrac{8 \pi^3}{6 \times 27}i = \dfrac{4 \pi^3}{81}i$$
We also have that $$\text{Li}_3(\omega) - \text{Li}_3(\omega^2) = (\omega-\omega^2)(f-g) = \sqrt{3}i(f-g)$$
Hence, we get that
$$f-g = \dfrac{4 \pi^3}{81 \sqrt3}$$
We can even get the values of $f$ and $g$ in terms of $\zeta(3)$. Note that $$f+g + \dfrac{\zeta(3)}{27} = \zeta(3) = \text{Li}_3(1)$$
Hence,
$$f = \dfrac{13}{27} \zeta(3) + \dfrac{2 \pi^3}{81 \sqrt3}; \,\,\,\,\,\,\,\, g = \dfrac{13}{27} \zeta(3) - \dfrac{2 \pi^3}{81 \sqrt3}$$
A: $\newcommand{\+}{^{\dagger}}
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${\tt\large\mbox{Just another one:}}$

\begin{align}
&\color{#66f}{\large\int_{0}^{1}{\ln^{2}\pars{x} \over 1 + x + x^{2}}\,\dd x}
=\lim_{\mu \to 0}\,\partiald[2]{}{\mu}
\int_{0}^{1}{\pars{1 - x}x^{\mu} \over 1 - x^{3}}\,\dd x
\\[3mm]&=\lim_{\mu \to 0}\,\partiald[2]{}{\mu}\int_{0}^{1}
{x^{\mu/3} - x^{\pars{\mu + 1}/3} \over 1 - x}\,{1 \over 3}\,x^{-2/3}\dd x
={1 \over 3}\,\lim_{\mu \to 0}\,\partiald[2]{}{\mu}\int_{0}^{1}
{x^{\pars{\mu - 2}/3} - x^{\pars{\mu - 1}/3} \over 1 - x}\,\dd x
\\[3mm]&={1 \over 3}\,\lim_{\mu \to 0}\,\partiald[2]{}{\mu}\bracks{%
\int_{0}^{1}{1 - x^{\pars{\mu - 1}/3} \over 1 - x}\,\dd x
-\int_{0}^{1}{1 - x^{\pars{\mu - 2}/3} \over 1 - x}\,\dd x}
\\[3mm]&={1 \over 3}\,\lim_{\mu \to 0}\,\partiald[2]{}{\mu}
\bracks{\Psi\pars{\mu + 2 \over 3} - \Psi\pars{\mu + 1 \over 3}}
={1 \over 27}\bracks{\Psi''\pars{2 \over 3} - \Psi''\pars{1 \over 3}}
\\[3mm]&={1 \over 27}\,\bracks{\pi\,\totald[2]{\cot\pars{\pi z}}{z}}_{z\ =\ ^1/_3}\
=\
{2\pi^{3} \over 27}\,\color{#c00000}{\cot\pars{\pi \over 3}}
\color{#f0f}{\csc^{2}\pars{\pi \over 3}}
={2\pi^{3} \over 27}\,\color{#c00000}{{1 \over \root{3}}}\,\color{#f0f}{\pars{2 \over \root{3}}^{2}}
\\[3mm]&=\color{#66f}{\large{8\root{3} \over 243}\,\pi^{3}}
\approx 1.7680
\end{align}
A: Rewrite
\begin{align}
\int_0^1 \dfrac{\ln^2 x }{1+x+x^2}\ dx&= \int_0^1 \dfrac{1-x}{1-x^3} \ln^2x\ dx\\
& = \int_0^1\sum_{k=0}^{\infty} x^{3k}\  (1-x) \ln^2x\ dx\\
&=\sum_{k=0}^{\infty}\left(\int_0^1 x^{3k}\ln^2x\ dx-\int_0^1 x^{3k+1}\ln^2x\ dx\right).\tag1
\end{align}
Using
$$
\int_0^1 x^\alpha \ln^n x\ dx=\frac{(-1)^n n!}{(\alpha+1)^{n+1}}, \qquad\text{for }\  n=0,1,2,\ldots\tag2
$$
then $(1)$ turns out to be
\begin{align}
\int_0^1 \dfrac{\ln^2 x }{1+x+x^2}\ dx&= \sum_{k=0}^{\infty}\left[\frac{2}{(3k+1)^{3}}-\frac{2}{(3k+2)^{3}}\right]\\
&= \frac1{3^3}\sum_{k=0}^{\infty}\left[\frac{2}{\left(k+\frac13\right)^{3}}-\frac{2}{\left(k+\frac23\right)^{3}}\right].\tag3\\
\end{align}
Consider series representation of polygamma function
$$
\psi_n(z)=(-1)^{n+1}\sum_{k=0}^\infty\frac{n!}{(k+z)^{n+1}}\quad;\quad\text{for}\ n>0\tag4
$$
and its reflection formula
$$
\psi_n(1-z)+(-1)^{n+1}\psi_n(z)=(-1)^{n}\pi\frac{d^n}{dz^n}\cot(\pi z).\tag5
$$
Using $(4)$ and $(5)$, then $(3)$ becomes
\begin{align}
\int_0^1 \dfrac{\ln^2 x }{1+x+x^2}\ dx&= \frac1{3^3}\bigg[\psi_2\left(\frac23\right)-\psi_2\left(\frac13\right)\bigg]\\
&=\frac\pi{27}\left.\frac{d^2}{dz^2}\cot(\pi z)\right|_{z=\frac13}\\
&=\large\color{blue}{\frac{8\sqrt3}{243}\pi^3}.
\end{align}
