Does $\int_{0}^1\frac{\ln(1+x+x^2)}{x}\mathrm dx$ have a closed form? 
$$\int_0^1 \frac{\ln(1+x+x^2)}{x} \mathrm{d}x = 1.09662$$

I am trying to find a closed form of this integral.
I think one might exist, this integral looks like it might be related to $\pi$, but I don't know.
 A: First
$$1+x+x^2=\frac{1-x^3}{1-x}$$
So rewrite the integrand as $$\int_{0}^{1} \frac{\ln(1-x^3)}{x}-\frac{\ln(1-x)}{x} dx.$$
But using u-subtitution $u=x^3,du=3x^2dx$
$$\int_{0}^{1} \frac{\ln(1-x^3)}{x}dx=\int_{0}^{1} \frac{\ln(1-u)}{3u}du$$,
so this means 
$$\int_{0}^{1} \frac{\ln(1-u)}{3u}du-\int_{0}^{1}\frac{\ln(1-x)}{x} dx=\frac{-2}{3}\int_{0}^{1}\frac{\ln(1-x)}{x} dx.$$
Now it is well known from the Basel Problem: 
$$\int_{0}^{1}\frac{\ln(1-x)}{x} dx=\int_{0}^{1}\int_{0}^{1}\frac{-1}{1-xy} dydx=-\zeta(2)=\frac{-\pi^2}{6}.$$
So $$\int_{0}^{1}\frac{\ln(1+x+x^2)}{x}dx=\frac{\pi^2}{9}$$
A: Note that $\ln(1+x+x^2)=\ln(1-x^3)-\ln(1-x)=-\sum_{k=1}^{\infty}(x^{3n}/n)+\sum_{k=1}^{\infty}(x^{n}/n)$. Change the order of integration and summation (it is valid, why?), and we have
$$\begin{align}\int_0^1 \frac{\ln(1+x+x^2)}{x} \mathrm{d}x&=\int_0^1\left(-\sum_{k=1}^\infty\frac{x^{3n-1}}{n}+\sum_{k=1}^\infty\frac{x^{n-1}}{n}\right)\mathrm{d}x\\
&=-\sum_{k=1}^\infty\int_0^1\frac{x^{3n-1}}{n}\mathrm dx+\sum_{k=1}^\infty\int_0^1\frac{x^{n-1}}{n}\mathrm dx\\
&=-\sum_{k=1}^\infty\frac{1}{3n^2}+\sum_{k=1}^\infty\frac{1}{n^2}=\frac{\pi^2}9.
\end{align}$$
