Integrate $\int_0^2 \frac{\ln\left(1+x\right)}{x^2-x+1} \mathop{dx}$ Challenge problem $$\int_0^2 \frac{\ln\left(1+x\right)}{x^2-x+1} \mathop{dx}$$
First thought $u=1+x$, $$ \int_1^3 \frac{\ln{(u)}}{u^2-3u+3} \mathop{du}$$
Here complex analysis or what?
Tips please.

Via the substitution $u=\sqrt{3}v$ we
get
\begin{equation*}
 I=\int_{1}^{3}\dfrac{\ln(u)}{u^2-3u+3}\,\mathrm{d}u =I_1+I_2
\end{equation*}
where
\begin{equation*}
 I_1=\dfrac{1}{\sqrt{3}}\int_{\frac{1}{\sqrt{3}}}^{\sqrt{3}}\dfrac{\ln(\sqrt{3})}{v^2-\sqrt{3}v+1}\,\mathrm{d}v =\left[\dfrac{\ln 3}{\sqrt{3}}\arctan\left(2v-\sqrt{3}\right)\right]_{\frac{1}{\sqrt{3}}}^{\sqrt{3}}=\dfrac{\pi\ln 3}{2\sqrt{3}}
\end{equation*}
and
\begin{equation*}
 I_2=\dfrac{1}{\sqrt{3}}\int_{\frac{1}{\sqrt{3}}}^{\sqrt{3}}\dfrac{\ln v}{v^2-\sqrt{3}v+1}\,\mathrm{d}v =[v\mapsto 1/v] =-I_2.
\end{equation*}
Consequently $I_2=0$ and
\begin{equation*}
 I=\dfrac{\pi\ln 3}{2\sqrt{3}}.
\end{equation*}
