Evaluating improper integral $\int_0^1 \frac{\log(x)}{x+\alpha}\; dx$ for small positive $\alpha$ Let $\alpha$ be a small positive real number.
How do I obtain
$$ I = \int_0^1 \frac{\log(x)}{x+\alpha}\; dx = -\frac{1}{2}(\log\alpha)^2 - \frac{\pi^2}{6} - \operatorname{Li}_2(-\alpha)$$? Maxima told me the result, but I do not understand how to get it.
The appearance of $\pi^2/6$ suggests the necessity of complex analysis. I tried to relate it to some contour integral through the change of variables $x=e^{-t}$:
$$ I = -\int_{0}^{\infty} \frac{t e^{-t}}{e^{-t}+\alpha}\;dt, $$
but I could not find the next step.

Assuming, for $u\leq 1$,
\begin{align}\text{Li}_2\left(u\right)&=-\int_0^1 \frac{u\ln x}{1-ux}dx\\
\text{Li}_2\left(1\right)&=\zeta(2)=\frac{\pi^2}{6}
\end{align}
Therefore, $0<\alpha<1$
\begin{align}\int_0^1 \frac{\log(x)}{x+\alpha}\; dx+\text{Li}_2\left(-\alpha\right)&=\underbrace{\frac{1}{\alpha}\int_0^1 \frac{\log(x)}{1+\frac{x}{\alpha}}\; dx}_{u=\frac{x}{\alpha}}+\underbrace{\int_0^1 \frac{\alpha\log(x)}{1+\alpha x}\; dx}_{u=\alpha x}\\
&=\int_0^{\frac{1}{\alpha}}\frac{\ln(\alpha u)}{1+u}du+\int_0^{\alpha}\frac{\ln(\frac{u}{\alpha} )}{1+u}du\\
&=\int_0^{\frac{1}{\alpha}}\frac{\ln u}{1+u}du+\int_0^{\alpha}\frac{\ln u}{1+u}du-\ln^2 \alpha\\
&=\left(\int_0^1 \frac{\ln x}{1+x}dx+\underbrace{\int_1^{\frac{1}{\alpha}} \frac{\ln x}{1+x}dx}_{u=\frac{1}{x}}\right)+\left(\int_0^1 \frac{\ln x}{1+x}dx-\int_\alpha^1 \frac{\ln x}{1+x}dx\right)-\\&\ln^2 \alpha\\
&=2\int_0^1 \frac{\ln x}{1+x}dx-\int_\alpha^1\frac{\ln x}{x(1+x)}dx-\int_\alpha^1 \frac{\ln x}{1+x}dx-\ln^2\alpha\\
&=2\int_0^1 \frac{\ln x}{1+x}dx+\int_\alpha^1\frac{\ln x}{1+x}dx-\underbrace{\int_\alpha^1\frac{\ln x}{x}dx}_{=-\frac{1}{2}\ln^2\alpha}-\int_\alpha^1 \frac{\ln x}{1+x}dx-\ln^2\alpha\\
&=2\int_0^1 \frac{\ln x}{1+x}dx-\frac{1}{2}\ln^2\alpha\\
&=2\left(\int_0^1 \frac{\ln x}{1-x}dx-\underbrace{\int_0^1 \frac{2x\ln x}{1-x^2}dx}_{u=x^2}\right)-\frac{1}{2}\ln^2\alpha\\
&=2\left(\int_0^1 \frac{\ln x}{1-x}dx-\frac{1}{2}\int_0^1 \frac{\ln u}{1-u}du\right)-\frac{1}{2}\ln^2\alpha\\
&=\int_0^1 \frac{\ln x}{1-x}dx-\frac{1}{2}\ln^2\alpha\\
&=-\text{Li}_2(1)-\frac{1}{2}\ln^2\alpha\\
&=-\frac{\pi^2}{6}-\frac{1}{2}\ln^2\alpha\\
\end{align}
