TITLE: Prove that $\int_{0}^{1} \ln\left(\frac{1-a x}{1-a}\right) \frac{1}{\ln x} \mathrm{dx} = -\sum_{k=1}^{\infty} a^{k} \frac{\ln(1+k)}{k}, \space a<1$
QUESTION [6 upvotes]: Prove that
$$\int_{0}^{1} \ln\left(\frac{1-a x}{1-a}\right) \frac{1}{\ln x} \mathrm{dx} =   -\sum_{k=1}^{\infty} a^{k} \frac{\ln(1+k)}{k}, \space a<1$$
I find this question rather troublesome since the both sides seem hard  to compute.
I appreciate any hint/suggestion. Thanks!

REPLY [3 votes]: Here's another way to handle the integral appearing in @Sasha's answer. 
$$\begin{eqnarray*}
\int_0^1 dx \, \frac{1-x^k}{\log x} &=& 
\int_\infty^{0} d\alpha\, \frac{\partial}{\partial \alpha}
\int_0^1 dx \, \frac{1-x^k}{\log x} x^\alpha \\
&=& \int_\infty^{0} d\alpha\, \int_0^1 dx \, (1-x^k)x^\alpha \\
&=& \int_\infty^{0} d\alpha\, \left(\frac{1}{\alpha+1} - \frac{1}{k+\alpha+1}\right) \\
&=& \left.\log\frac{\alpha+1}{k+\alpha+1}\right|_\infty^{0} \\
&=& -\log(k+1)
\end{eqnarray*}$$