Seeking alternative methods for $\int _0^1\frac{\ln \left(x^2-x+1\right)}{x\left(1-x\right)}\:dx$ I've solved this one by first tackling,
$$\int _0^{\infty }\frac{\ln \left(x^2-x+1\right)}{x\left(1-x\right)}\:dx$$
But i'd like to know other ways to solve it since the way i did it was a bit lengthy and not that straightforward.
 A: $$\int_0^1\frac{\ln(x^2-x+1)}{x(1-x)}\ dx=\underbrace{\int_0^1\frac{\ln(x^2-x+1)}{1-x}\ dx}_{x\to 1-x}+\int_0^1\frac{\ln(x^2-x+1)}{x}\ dx$$
$$=2\int_0^1\frac{\ln(x^2-x+1)}{x}\ dx=2\underbrace{\int_0^1\frac{\ln(x^3+1)}{x}\ dx}_{x^3\to x}-2\int_0^1\frac{\ln(1+x)}{x}\ dx$$
$$=-\frac43\int_0^1\frac{\ln(1+x)}{x}\ dx=-\frac43\cdot\frac12\zeta(2)=-\frac{\pi^2}{9}$$

A different way to calculate the last integral is to use the identity
$$\sum_{n=1}^{\infty}\frac{x^n}{n}\cos(an)=-\frac12\ln(1-2x\cos(a)+x^2)$$
Set $a=\frac{\pi}{3}$ we get
$$\ln(1-x+x^2)=-2\sum_{n=1}^\infty \frac{x^n}{n}\cos(n\pi/3)$$
so
$$\int_0^1\frac{\ln(1-x+x^2)}{x}\ dx=-2\sum_{n=1}^\infty \frac{\cos(n\pi/3)}{n}\int_0^1 x^{n-1}\ dx$$
$$=2\sum_{n=1}^\infty\frac{\cos(n\pi/3)}{n^2}=-\frac{\pi^2}{18}$$
where the last result follows from integrating both sides of the common identity
$$\sum_{n=1}^\infty \frac{\sin(nx)}{n}=\frac{\pi-x}{2}$$
A: My approach.
$$\int _0^{\infty }\frac{\ln \left(x^2-x+1\right)}{x\left(1-x\right)}\:dx$$
$$=\int _0^1\frac{\ln \left(x^2-x+1\right)}{x\left(1-x\right)}\:dx\:+\int _1^{\infty }\frac{\ln \left(x^2-x+1\right)}{x\left(1-x\right)}\:dx$$
Let $\displaystyle x=\frac{1}{t}$ for the $2$nd integral.
$$\int _0^1\frac{\ln \left(t^2-t+1\right)}{t\left(1-t\right)}\:dt\:+\int _0^1\frac{\ln \left(t^2-t+1\right)}{t-1}\:dt-2\int _0^1\frac{\ln \left(t\right)}{t-1}\:dt$$
$$=\int _0^1\left(\frac{1}{t\left(1-t\right)}+\frac{1}{t-1}\right)\ln \left(t^2-t+1\right)\:dt\:-2\sum _{k=0}^{\infty }\frac{1}{\left(k+1\right)^2}$$
$$=\int _0^1\frac{\ln \left(t^2-t+1\right)}{t}\:dt\:-\frac{\pi ^2}{3}$$
$$\int _0^1\frac{\ln \left(t^3+1\right)}{t}\:dt-\int _0^1\frac{\ln \left(t+1\right)}{t}\:dt-\frac{\pi ^2}{3}$$
$$\int _0^1\sum _{k=1}^{\infty }\frac{\left(-1\right)^{k+1}}{k}t^{3k-1}\:dt-\int _0^1\sum _{k=1}^{\infty }\frac{\left(-1\right)^{k+1}}{k}t^{k-1}\:dt-\frac{\pi ^2}{3}$$
$$\sum _{k=1}^{\infty }\frac{\left(-1\right)^{k+1}}{3k^2}+\sum _{k=1}^{\infty }\frac{\left(-1\right)^{k+2}}{k^2}-\frac{\pi ^2}{3}$$
$$\frac{\pi ^2}{36}-\frac{\pi ^2}{12}-\frac{\pi ^2}{3}=-\frac{7\pi ^2}{18}$$
So,
$$\boxed{\int _0^{\infty }\frac{\ln \left(x^2-x+1\right)}{x\left(1-x\right)}\:dx=-\frac{7\pi ^2}{18}}$$
In order to find the desired integral i used this previous expression.
$$\int _0^{\infty }\frac{\ln \left(x^2-x+1\right)}{x\left(1-x\right)}\:dx=\int _0^1\frac{\ln \left(t^2-t+1\right)}{t}\:dt\:-\frac{\pi ^2}{3}$$
And let $\displaystyle t=1-u$
$$-\frac{7\pi ^2}{18}=-\int _0^1\frac{\ln \left(u^2-u+1\right)}{u-1}\:du\:-\frac{\pi ^2}{3}$$
$$\boxed{\int _0^1\frac{\ln \left(u^2-u+1\right)}{u-1}\:du\:=\frac{\pi ^2}{18}}$$
Notice that on the $3$rd line all we have to do is to put in the result we just found and we'll be done.
$$-\frac{7\pi ^2}{18}=\int _0^1\frac{\ln \:\left(t^2-t+1\right)}{t\left(1-t\right)}\:dt\:+\frac{\pi ^2}{18}-\frac{\pi ^2}{3}$$
And finally.
$$\boxed{\int _0^1\frac{\ln \:\left(t^2-t+1\right)}{t\left(1-t\right)}\:dt\:=-\frac{\pi ^2}{9}}$$
A: On the path of Dennis Orton...
\begin{align}J&=\int _0^1\frac{\ln \left(x^2-x+1\right)}{x\left(1-x\right)}\:dx\\
&=\int _0^1\frac{\ln \left(x^2-x+1\right)}{x}\:dx+\int _0^1\frac{\ln \left(t^2-t+1\right)}{1-t}\:dt\\
&\overset{x=1-t}=2\int _0^1\frac{\ln \left(x^2-x+1\right)}{x}\:dx\\
&=2\left(\int _0^1\frac{\ln \left(\frac{1+x^3}{1+x}\right)}{x}\:dx\right)\\
&=2\left(\int _0^1\frac{\ln \left(1+t^3\right)}{t}\:dt-\int _0^1\frac{\ln \left(1+x\right)}{x}\:dx\right)\\
&\overset{x=t^3}=\frac{2}{3}\int _0^1\frac{\ln \left(1+x\right)}{x}\:dt-2\int _0^1\frac{\ln \left(1+x\right)}{x}\:dx\\
&=-\frac{4}{3}\int _0^1\frac{\ln \left(1+x\right)}{x}\:dx\\
&=-\frac{4}{3}\left(\int_0^1\frac{\ln \left(1-t^2\right)}{t}\:dt-\int _0^1\frac{\ln \left(1-x\right)}{x}\:dx\right)\\
&\overset{x=t^2}=-\frac{4}{3}\left(\frac{1}{2} \int _0^1\frac{\ln \left(1-x\right)}{x}\:dx-\int _0^1\frac{\ln \left(1-x\right)}{x}\:dx\right)\\
&=\frac{2}{3}\int _0^1\frac{\ln \left(1-x\right)}{x}\:dx\\
&=\frac{2}{3}\left(-\int_0^1 \left(\sum_{n=1}^\infty\frac{x^{n-1}}{n}\right)\,dx\right)\\
&=-\frac{2}{3}\sum_{n=1}^\infty\left(\int_0^1 \frac{x^{n-1}}{n}\,dx\right)\\
&=-\frac{2}{3}\sum_{n=1}^\infty\frac{1}{n^2}\\
&=-\frac{2}{3}\times\frac{\pi^2}{6}\\
&=\boxed{-\frac{\pi^2}{9}}
\end{align}
NB: I assume $\displaystyle \sum_{n=1}^\infty\frac{1}{n^2}=\zeta(2)=\frac{\pi^2}{6}$
PS:
Sorry, i didn't see the soluton of Ali Shather
A: I don't know if you like this or not. Let
$$I(\alpha)=\int _0^1\frac{\ln \left[4\sin^2(\alpha)(x^2-x)+1\right]}{x\left(1-x\right)}\:dx, \alpha\in[0,\pi/6]$$
Then $I(0)=0, I(\frac{\pi}{6})=I$. Since
\begin{eqnarray}
I'(\alpha)&=&-\int _0^1\frac{4\sin(2\alpha)}{4\sin^2(\alpha)(x^2-x)+1}\,dx\\
&=&-\int _0^1\frac{2\cot(\alpha)}{x^2-x+\frac1{4\sin^2(\alpha)}}\,dx\\
&=&-\int _0^1\frac{2\cot(\alpha)}{(x-\frac12)^2+\frac1{4}\cot^2(\alpha)}\,dx\\
&=&-8\alpha.
\end{eqnarray}
So
$$ I=\int_0^{\pi/6}I'(\alpha)\;d\alpha=-\int_0^{\pi/6}8\alpha\;d\alpha=-\frac{\pi^2}{9}. $$
A: Expanding in series and integrating gives
$$\newcommand{\Li}{\operatorname{Li}}
\int_0^1\log(1+\alpha x)\,\frac{\mathrm{d}x}x=-\Li_2(-\alpha)\tag1
$$
Setting $\omega=e^{i2\pi/3}$, we get
$$
\begin{align}
\int_0^1\frac{\log\left(x^2-x+1\right)}{x(1-x)}\,\mathrm{d}x
&=\int_0^1\log\left(1-x+x^2\right)\left(\frac1x+\frac1{1-x}\right)\mathrm{d}x\tag2\\
&=2\int_0^1\log\left(1-x+x^2\right)\frac{\mathrm{d}x}x\tag3\\
&=2\int_0^1\left(\log\left(1+\omega x\right)+\log\left(1+\omega^2x\right)\right)\frac{\mathrm{d}x}x\tag4\\[6pt]
&=-2\left(\Li(e^{i\pi/3})+\Li\left(e^{-i\pi/3}\right)\right)\tag5\\[6pt]
&=-2\left(\frac{\pi^2}3-\frac{5\pi^2}{18}\right)\tag6\\
&=-\frac{\pi^2}9\tag7
\end{align}
$$
Explanation:
$(2)$: partial fractions
$(3)$: substitute $x\mapsto1-x$ to get $\frac1{1-x}\mapsto\frac1x$
$(4)$: factor $1-x+x^2$
$(5)$: apply $(1)$
$(6)$: apply $(14)$ from this answer
$(7)$: simplfy
A: Here is another way to do it :
Let's start by using the substitution $ \left\lbrace\begin{aligned}t&=\frac{1-\sqrt{1-x}}{2}\\ \mathrm{d}t&=\frac{\mathrm{d}x}{4\sqrt{1-x}}\end{aligned}\right. $, we have : $$ \int_{0}^{\frac{1}{2}}{\frac{\ln{\left(1-t+t^{2}\right)}}{t\left(1-t\right)}\,\mathrm{d}x}=\int_{0}^{1}{\frac{\ln{\left(1-\frac{x}{4}\right)}}{x\sqrt{1-x}}\,\mathrm{d}x} $$
By using the substitution $ \left\lbrace\begin{aligned}t&=\frac{1+\sqrt{1-x}}{2}\\ \mathrm{d}t&=-\frac{\mathrm{d}x}{4\sqrt{1-x}}\end{aligned}\right. $, we have : $$ \int_{\frac{1}{2}}^{1}{\frac{\ln{\left(1-t+t^{2}\right)}}{t\left(1-t\right)}\,\mathrm{d}x}=\int_{0}^{1}{\frac{\ln{\left(1-\frac{x}{4}\right)}}{x\sqrt{1-x}}\,\mathrm{d}x} $$
Thus : $$ \int_{0}^{1}{\frac{\ln{\left(1-t+t^{2}\right)}}{t\left(1-t\right)}\,\mathrm{d}x}=2\int_{0}^{1}{\frac{\ln{\left(1-\frac{x}{4}\right)}}{x\sqrt{1-x}}\,\mathrm{d}x} $$
Let's work now on $ \int_{0}^{1}{\frac{\ln{\left(1-\frac{x}{4}\right)}}{x\sqrt{1-x}}\,\mathrm{d}x} $ shall we :
Substitute : $ \left\lbrace\begin{aligned} u &=\sqrt{1-x} \\ \mathrm{d}u &=-\frac{\mathrm{d}x}{2\sqrt{1-x}} \end{aligned}\right. $, we get :
$ \displaystyle\int_{0}^{1}{\displaystyle\frac{\ln{\left(1-\frac{x}{4}\right)}}{x\sqrt{1-x}}\,\mathrm{d}x}=2\displaystyle\int_{0}^{1}{\displaystyle\frac{\ln{\left(\frac{3+u^{2}}{4}\right)}}{1-u^{2}}\,\mathrm{d}u}$
Since : $ \left(\forall u\in\left[0,1\right]\right),\ \displaystyle\int_{0}^{1}{\displaystyle\frac{1-u^{2}}{\left(1-u^{2}\right)v-4}\,\mathrm{d}v}=\ln{\left(\displaystyle\frac{3+u^{2}}{4}\right)}$, we have :
\begin{aligned}\displaystyle\int_{0}^{1}{\displaystyle\frac{\ln{\left(1-\frac{x}{4}\right)}}{x\sqrt{1-x}}\,\mathrm{d}x}&=-2\displaystyle\int_{0}^{1}\displaystyle\int_{0}^{1}{\displaystyle\frac{\mathrm{d}v\,\mathrm{d}u}{v u^{2}+4-v}}\\&=-2\displaystyle\int_{0}^{1}{\displaystyle\int_{0}^{1}{\displaystyle\frac{\mathrm{d}u}{v u^{2}+4-v}}\,\mathrm{d}v}\\&=-2\displaystyle\int_{0}^{1}{\displaystyle\frac{1}{\sqrt{v}\sqrt{4-v}}\displaystyle\int_{0}^{1}{\displaystyle\frac{\sqrt{\frac{v}{4-v}}}{1+\left(\sqrt{\frac{v}{4-v}}u\right)^{2}}\,\mathrm{d}u}\,\mathrm{d}v}\\ \displaystyle\int_{0}^{1}{\displaystyle\frac{\ln{\left(1-\frac{x}{4}\right)}}{x\sqrt{1-x}}\,\mathrm{d}x}&=-2\displaystyle\int_{0}^{1}{\displaystyle\frac{1}{\sqrt{v}\sqrt{4-v}}\arctan{\left(\sqrt{\frac{v}{4-v}}\right)}\,\mathrm{d}v}\end{aligned}
With the substitution : $ \left\lbrace\begin{aligned}\alpha &=\frac{\sqrt{v}}{2} \\ \mathrm{d}\alpha &=\displaystyle\frac{\mathrm{d}v}{4\sqrt{v}}\end{aligned}\right. $, and the fact that $ \left(\forall x\in\left]-1,1\right[\right),\ \arctan{\left(\displaystyle\frac{x}{\sqrt{1-x^{2}}}\right)}=\arcsin{x} $, we get :
$ \displaystyle\int_{0}^{1}{\displaystyle\frac{\ln{\left(1-\frac{x}{4}\right)}}{x\sqrt{1-x}}\,\mathrm{d}x}=-4\displaystyle\int_{0}^{\frac{1}{2}}{\displaystyle\frac{\arcsin{\alpha}}{\sqrt{1-\alpha^{2}}}\,\mathrm{d}\alpha}=-2\left[\arcsin^{2}{\alpha}\right]_{0}^{\frac{1}{2}}=-\displaystyle\frac{\pi^{2}}{18} \cdot $
Thus : $$ \int_{0}^{1}{\frac{\ln{\left(1-t+t^{2}\right)}}{t\left(1-t\right)}\,\mathrm{d}x}=-\frac{\pi^{2}}{9} $$
A: Note $ \frac1{x(1-x)} = \frac1x+\frac1{1-x}$ to obtain
\begin{align}
I=\int_0^1\frac{\ln\left(x^2-x+1\right)}{x(1-x)}dx
&=2\int_0^1\frac{dx}x  \ln\left(1-x+x^2\right)
\end{align}
Let $J(a)=\int_0^1\frac{dx}x\ln\left(1-2\cos a x+x^2\right)$
$$J’(a)=\int_0^1 \frac{2\sin a}{(x-\sin a)^2+\sin^2a}dx=\pi-a
$$
Note
$$J(\frac\pi2)= \int_0^1\frac{\ln(1+x^2)}{x}dx\overset{x^2\to x}=\frac12\int_0^1\frac{\ln(1+x)}{x}dx\\
= \frac12\int_0^1\frac{\ln(1+x^3)-\ln(1-x+x^2)}{x}dx
=\frac16\int_0^1\frac{\ln(1+x)}{x}dx-\frac14I =-\frac38 I
$$
Then
\begin{align}
I=2J(\frac\pi3) = 2\left(J(\frac\pi2)- \int_{\pi/3}^{\pi/2}J’(a)da\right)=-\frac34I -2 \int_{\pi/3}^{\pi/2}(\pi-a)da
\end{align}
which leads to $I= -\frac{\pi^2}9$.
