Evaluating $\int _0^1\frac{\ln \left(x^2+x+1\right)}{x\left(x+1\right)}\:dx$ How can i evaluate this integral, maybe differentiation under the integral sign? i started expressing the integral as the following,
$$\int _0^1\frac{\ln \left(x^2+x+1\right)}{x\left(x+1\right)}\:dx=\int _0^1\frac{\ln \left(x^2+x+1\right)}{x}\:dx-\int _0^1\frac{\ln \left(x^2+x+1\right)}{x+1}\:dx\:$$
But i dont know how to keep going, ill appreciate any solutions or hints.
 A: I dont think Feynman's trick would work best here, following your path:
$$\int _0^1\frac{\ln \left(x^2+x+1\right)}{x\left(x+1\right)}\:dx=\int _0^1\frac{\ln \left(x^2+x+1\right)}{x}\:dx-\underbrace{\int _0^1\frac{\ln \left(x^2+x+1\right)}{x+1}\:dx}_{x=\frac{1-t}{1+t}}\:$$
$$=\int _0^1\frac{\ln \left(x^3-1\right)}{x}\:dx-\int _0^1\frac{\ln \left(x-1\right)}{x}\:dx-\int _0^1\frac{\ln \left(x^2+3\right)}{x+1}\:dx+2\int _0^1\frac{\ln \left(x+1\right)}{x+1}\:dx$$
$$-\sum _{k=1}^{\infty }\frac{1}{k}\int _0^1x^{3k-1}\:dx\:+\sum _{k=1}^{\infty }\frac{1}{k}\:\int _0^1x^{k-1}\:dx-\int _0^1\frac{\ln \left(x^2+3\right)}{x+1}\:dx+\ln ^2\left(2\right)$$
To solve that remaining integral you can use the identity i derived here
So,
$$=\frac{2\zeta \left(2\right)}{3}-(-\frac{\ln ^2\left(3\right)}{4}-\frac{\text{Li}_2\left(-\frac{1}{3}\right)}{2}-\frac{\ln ^2\left(4\right)}{4}+\frac{\ln \left(3\right)\ln \left(4\right)}{2}-\arctan ^2\left(\sqrt{\frac{1}{3}}\right)+\ln \left(2\right)\ln \left(4\right))+\ln ^2\left(2\right)$$
$$\frac{\pi ^2}{9}+\frac{\ln ^2\left(3\right)}{4}+\frac{\text{Li}_2\left(-\frac{1}{3}\right)}{2}+\ln ^2\left(2\right)-\ln \left(3\right)\ln \left(2\right)+\frac{\pi ^2}{36}-2\ln ^2\left(2\right)+\ln ^2\left(2\right)$$
So your integral's solution is,
$$\boxed{\int _0^1\frac{\ln \left(x^2+x+1\right)}{x\left(x+1\right)}\:dx=\frac{5\pi ^2}{36}+\frac{\ln ^2\left(3\right)}{4}+\frac{\text{Li}_2\left(-\frac{1}{3}\right)}{2}-\ln \left(3\right)\ln \left(2\right)}$$
A: Solution using harmonic series
$$\int _0^1\frac{\ln \left(x^2+x+1\right)}{x\left(x+1\right)}\:dx=\int _0^1\frac{\ln \left(x^2+x+1\right)}{x}\:dx-\int _0^1\frac{\ln \left(x^2+x+1\right)}{x+1}\:dx\:$$

$$\int _0^1\frac{\ln \left(x^2+x+1\right)}{x}\:dx=\underbrace{\int _0^1\frac{\ln \left(1-x^3\right)}{x}\:dx}_{x^3\to x}-\int _0^1\frac{\ln \left(1-x\right)}{x}\:dx$$
$$=-\frac23\int _0^1\frac{\ln \left(1-x\right)}{x}\:dx=\frac23\zeta(2)$$

$$\int _0^1\frac{\ln \left(1+x+x^2\right)}{1+x}\:dx\overset{IBP}{=}\ln(2)\ln(3)-\int_0^1\frac{(2x+1)\ln(1+x)}{1+x+x^2}dx$$
For the latter integral, set $a=\frac{2\pi}{3}$ in the identity
$$\sum_{n=1}^{\infty}x^{n-1} \cos(na)=\frac{\cos(a)-x}{1-2x\cos(a)+x^2}, \ |x|<1$$
we have
$$-2\sum_{n=1}^{\infty}x^{n-1} \cos(n\frac{2\pi}{3})=\frac{2x+1}{1+x+x^2}$$
$$\Longrightarrow \int_0^1\frac{(2x+1)\ln(1+x)}{1+x+x^2}dx=-2\sum_{n=1}^\infty \cos(n\frac{2\pi}{3})\int_0^1 x^{n-1}\ln(1+x)dx$$
$$=-2\sum_{n=1}^\infty \cos(n\frac{2\pi}{3})\left(\frac{H_n-H_{n/2}}{n}\right)$$
$$=-2\Re\sum_{n=1}^\infty \left(e^{i\frac{2\pi}{3}}\right)^n\left(\frac{H_n-H_{n/2}}{n}\right)$$
And finally we use the generating functions
$$\sum_{n=1}^\infty  x^n\frac{H_n}{n}=\frac12\ln^2(1-x)+\text{Li}_2(x)$$
$$\sum_{n=1}^\infty  x^n\frac{H_{n/2}}{n}=i\pi\frac{\ln(1-x^2)-\ln(-x^2)}{x^2}$$
$$+\frac{\ln(x-1)\ln(-x^2)-\ln(x-1)\ln(1-x^2)}{x^2}$$
$$+\frac{\text{Li}_2\left(\frac{1-x}{1+x}\right)-\text{Li}_2\left(\frac{1}{1+x}\right)-\text{Li}_2\left(\frac{1}{1-x}\right)}{x^2}$$
I found the second generating function with help of Mathematica after I converted it to integral;
$$\sum_{n=1}^\infty  x^n\frac{H_{n/2}}{n}=-\int_0^1\frac{xy^2\ln(1-y^2)}{1-xy}dy$$
