Prove that $\int_{0}^{1} \ln\left(\frac{1-a x}{1-a}\right) \frac{1}{\ln x} \mathrm{dx} = -\sum_{k=1}^{\infty} a^{k} \frac{\ln(1+k)}{k}, \space a<1$ 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!
 A: Develop series expansion of the logarithm of the quotient:
$$
   \ln \left( \frac{1-a x}{1-a} \right) = \ln\left(1-a x\right) - \ln\left(1-a\right) = \sum_{k=1}^\infty \frac{1}{k} a^k \left( 1-x^k \right)
$$
It is now down to evaluation of 
$$ \begin{eqnarray}
  \int_0^1 \frac{1-x^k}{\ln(x)} \mathrm{d}x &\stackrel{x = \exp(-t)}{=}&  -\int_0^\infty \mathrm{e}^{-t} \frac{1-\exp(-k t)}{t} \mathrm{d}t \\ &=& \int_0^{k} \frac{\mathrm{d}}{\mathrm{d}m } \left(-\int_0^\infty \mathrm{e}^{-t} \frac{1-\exp(-m t)}{t} \mathrm{d}t\right)\mathrm{d}m \\
  &=& -\int_0^k \int_0^\infty \exp(-t(m+1)) \mathrm{d}m \\ &=& -\int_0^k \frac{\mathrm{d}m}{m+1} = -\log(k+1)
\end{eqnarray}
$$ 
A: Here's another way to handle the integral appearing in @Sasha's answer. 
$$\begin{eqnarray*}
\int_0^1 dx \, \frac{1-x^k}{\log x} &=& 
\int_\infty^{0} d\alpha\, \frac{\partial}{\partial \alpha}
\int_0^1 dx \, \frac{1-x^k}{\log x} x^\alpha \\
&=& \int_\infty^{0} d\alpha\, \int_0^1 dx \, (1-x^k)x^\alpha \\
&=& \int_\infty^{0} d\alpha\, \left(\frac{1}{\alpha+1} - \frac{1}{k+\alpha+1}\right) \\
&=& \left.\log\frac{\alpha+1}{k+\alpha+1}\right|_\infty^{0} \\
&=& -\log(k+1)
\end{eqnarray*}$$
