<s>[INST] How to solve $\int_0^{2}\frac{x\log(2-x)}{1+x^2} \; dx $ I have problem to find the value of the integral:
$$ \int_0^{2}\frac{x\log(2-x)}{1+x^2} \; dx $$
The first step is to split the absolute part.
I thank you in advance for help.
 [/INST] Using
\begin{eqnarray} 
&&\int \frac{\log(b-x)}{x+a}\;dx\\
&=&\int \log(b-x)d\log\frac{x+a}{a+b}\;dx\\
&=&\log(b-x)\log\frac{x+a}{a+b}-\int \frac1{b-x}\log\frac{x+a}{a+b}\;dx\\
&=&\log(b-x)\log\frac{x+a}{a+b}-\int \frac1{b-x}\log\bigg(1-\frac{b-x}{a+b}\bigg)\;dx\\
&=&\log(b-x)\log\frac{x+a}{a+b}+\text{Li}_2(\frac{b-x}{a+b})+C
\end{eqnarray}
one has
\begin{eqnarray}
&&\int_0^{2}\frac{x\log(2-x)}{1+x^2} \; dx\\
&=&\frac12\int_0^2\log(2-x)\bigg(\frac1{x+i}+\frac1{x-i}\bigg)\,dx\\
&=&\frac12\log2\log5-\frac12\bigg[\text{Li}_2\bigg(\frac{4}{5}-\frac{2i}{5}\bigg)+\text{Li}_2\bigg(\frac{4}{5}+\frac{2i}{5}\bigg)\bigg]
\end{eqnarray}
</s>