How to evaluate $\int _0^1\frac{\arctan \left(\frac{x^2-x}{x-2}\right)}{x}\:dx$

Question

I want to know what are the best ways to evaluate $$\int _0^1\frac{\arctan \left(\frac{x^2-x}{x-2}\right)}{x}\:dx$$ It seems that it is equal to $\frac{G}{3}$ where $G$ is the Catalan's constant, I tried integration by parts and ended with $$-\int _0^1\frac{\ln \left(x\right)\left(2-4x+x^2\right)}{4-2x^3+2x^2-4x+x^4}\:dx$$ but this just looks complicated, I also tried to substitute the argument of the $\arctan$ and the sub $\frac{1-t}{1+t}$ but neither simplified things, can i please be given some hint on how to tackle this?.

The original integral also equals $$-\frac{i}{2}\left(\int _0^1\frac{\ln \left(-2i+ix+x-x^2\right)}{x}\:dx-\int _0^1\frac{\ln \left(-2i+ix-x+x^2\right)}{x}\:dx\right)$$ But I'd prefer to avoid complex methods.

Answer 1

Consider $$\int \frac{2-4x+x^2}{4-2x^3+2x^2-4x+x^4}\,\,\log \left(x\right)\:dx$$ The quartic in denominator shows four complex roots; let us call them $(a,b,c,d)$. Using partial fractions, we then end with four integrals looking like $$I_k=\int \frac{\log(x)}{x-k}\,dx=\text{Li}_2\left(\frac{x}{k}\right)+\log (x) \log \left(1-\frac{x}{k}\right)$$ $$J_k=\int_0^1 \frac{\log(x)}{x-k}\,dx=\text{Li}_2\left(\frac{1}{k}\right)$$ and the nightmare $$2\sqrt 2 \,\int_0^1 \frac{2-4x+x^2}{4-2x^3+2x^2-4x+x^4}\,\,\log \left(x\right)\:dx$$ $$-i \left(\sqrt{2} \text{Li}_2\left(\left(-\frac{1+i}{2}\right) (-1)^{1/6}\right)-\left(\sqrt{3}-1\right) \sqrt{2+\sqrt{3}} \text{Li}_2\left(\left(\frac{1-i}{2}\right) (-1)^{1/6}\right)+\sqrt{2} \left(\text{Li}_2\left(\left(\frac{1+i}{4}\right) \left(\sqrt{3}-i\right)\right)-\text{Li}_2\left(\left(\frac{1-i}{2} \right) (-1)^{5/6}\right)\right)\right)$$ since $$a=\frac{1}{2} \left(1+\sqrt{3}-i \sqrt{2 \left(2-\sqrt{3}\right)}\right)\qquad b=\frac{1}{2} \left(1+\sqrt{3}+i \sqrt{2 \left(2-\sqrt{3}\right)}\right)$$ $$c=\frac{1}{2} \left(1-\sqrt{3}-i \sqrt{2 \left(2+\sqrt{3}\right)}\right)\qquad d=\frac{1}{2} \left(1-\sqrt{3}+i \sqrt{2 \left(2+\sqrt{3}\right)}\right)$$ My problem is that I cannot simplify the end result (which numerically is exactly $-\frac{C}{3}$.