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.
 A: 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*}
A: $\mathcal{Hint:}$
Substitute $u=3/t$
By the way, there shouldn't be a $1/3$ outside of your integral.
A: As said in comments, the $\frac 13$ appeared without any reason. So the problem is simply
$$I= \int_1^3 \frac{\log{(u)}}{u^2-3u+3} \,du$$
Rewrite $$u^2-3u+3=(u-a)(u-b)\implies \frac 1{u^2-3u+3}=\frac 1{a-b} \left(\frac 1{u-a} -\frac 1{u-b} \right)$$ which makes that we face two integrals looking like
$$\int \frac{\log{(u)}}{u-c}\,du=\text{Li}_2\left(\frac{u}{c}\right)+\log (u) \log \left(1-\frac{u}{c}\right)$$ obtained after one integration by parts.
Using $a=\frac{3+i \sqrt{3}}{2} $ and $b=\frac{3-i \sqrt{3}}{2} $  this gives  the antiderivative.
Using the bounds and simplifying the polylogarithms leads to
$$I=\frac{\pi  \log (3)}{2 \sqrt{3}}$$
