Definite Integral $\int_0^2 \frac{\ln(1+x)}{x^2-x+1}dx$

Question

I was working on this problem, I probably need to abuse symmetry somewhere but can't see how:

$$\int_0^2 \frac{\ln(1+x)}{x^2-x+1}dx$$

Answer 1

\begin{align*} I&=\underbrace{\int_0^2\frac{\ln(x+1)}{x^2-x+1}dx}_{x+1\mapsto x}\\ &=\underbrace{\int_1^3\frac{\ln x}{x^2-3x+3}dx}_{x\mapsto 3/x}\\ &=\int_3^1\frac{\ln\left(\frac{3}{x}\right)}{\frac{9}{x^2}-\frac{9}{x}+3}\left(-\frac{3}{x^2}dx\right)\\ &=\int_1^3\frac{\ln 3-\ln x}{x^2-3x+3}dx\\ &=\int_1^3\frac{\ln 3}{x^2-3x+3}dx-I\\ &=\frac{\ln 3}{2}\int_1^3\frac{1}{x^2-3x+3}dx\\ &=\frac{\ln 3}{2}\left[\frac{2}{\sqrt{3}}\arctan\left(\frac{2}{\sqrt{3}}\left(x-\frac{3}{2}\right)\right)\right]_{x=1}^3\\ &=\frac{\ln 3}{\sqrt{3}}\left(\arctan\left(\sqrt{3}\right)-\arctan\left(-\frac{1}{\sqrt{3}}\right)\right)\\ &=\frac{\ln 3}{\sqrt{3}}\left(\frac{\pi}{3}+\frac{\pi}{6}\right)\\ &=\frac{\pi\ln 3}{2\sqrt{3}} \end{align*}

Answer 2

As $x^2-x+1=\dfrac{(2x-1)^2+3}4,$

set $2x-1=\sqrt3\tan t,2dx=\sqrt3\sec^2t\ dt$

$$I=\int_0^2\dfrac{\ln(1+x)}{x^2-x+1}dx=\int_{-\pi/6}^{\pi/3}\dfrac{\ln\left(1+\dfrac{\sqrt3\tan t+1}2\right)}{\dfrac{3\sec^2t}4}\cdot\dfrac{\sqrt3\sec^2t\ dt}2$$

Now $1+\dfrac{\sqrt3\tan t+1}2=\sqrt3\cdot\dfrac{\sqrt3\cos t+\sin t}{2\cos t}$

$$\implies\dfrac{\sqrt3I}2=\ln\sqrt3\int_{-\pi/6}^{\pi/3}dt+\int_{-\pi/6}^{\pi/3}\ln\cos\left(t-\dfrac\pi6\right)dt-\int_{-\pi/6}^{\pi/3}\ln\cos t\ dt$$

$$=\dfrac{\ln3}2\int_{-\pi/6}^{\pi/3}dt$$

as like Evaluate the integral $\int^{\frac{\pi}{2}}_0 \frac{\sin^3x}{\sin^3x+\cos^3x}\,\mathrm dx$.,

$$\int_{-\pi/6}^{\pi/3}\ln\cos\left(t-\dfrac\pi6\right)dt=\int_{-\pi/6}^{\pi/3}\ln\cos\left(\left(\dfrac\pi3-\dfrac\pi6-t\right)-\dfrac\pi6\right)dt=\int_{-\pi/6}^{\pi/3}\ln\cos t\ dt$$