How to compute the integral $ I\left(c\right)=\int_{0}^{1}{\frac{\ln(1-cx)}{1+x}dx} $ I am currently working on this question and the following integral came up:
$$
I\left(c\right)=\int_{0}^{1}{\frac{\ln(1-cx)}{1+x}dx}
$$
for a suitable c. I would like to compute it in terms of $\operatorname{Li}_2$. I tried to expand the logarithm, but things got a bit tedious. So any help is highly appreciated.
 A: 
I would like to offer a generalization to this problem, which turns out to be a useful lemma in more difficult logarithmic integral problems.
Define the function $\mathcal{D}:\left(-\infty,1\right)\times\left(-\infty,1\right]\rightarrow\mathbb{R}$ via the definite integral
$$\mathcal{D}{\left(a,b\right)}:=\int_{0}^{1}\mathrm{d}y\,\frac{a\ln{\left(1-by\right)}}{ay-1}.\tag{1}$$
We show that the integral $\mathcal{D}$ has the following closed-form expression in terms of dilogarithms:
$$\forall\left(a,b\right)\in\left(-\infty,1\right)\times\left(-\infty,1\right]:\mathcal{D}{\left(a,b\right)}=\operatorname{Li}_{2}{\left(\frac{a-b}{a-1}\right)}-\operatorname{Li}_{2}{\left(\frac{a}{a-1}\right)}-\operatorname{Li}_{2}{\left(b\right)}.\tag{2}$$

Proof:
It is easy to check that the RHS of $(2)$ yields the correct value of zero for $\mathcal{D}$ in special case where at least one of the parameters vanishes identically.
For the remaining general case where $\left(a,b\right)\in\left(-\infty,1\right)\times\left(-\infty,1\right]\land a\neq0\land b\neq0$, we obtain
$$\begin{align}
\mathcal{D}{\left(a,b\right)}
&=\int_{0}^{1}\mathrm{d}y\,\frac{a\ln{\left(1-by\right)}}{ay-1}\\
&=\int_{0}^{1}\mathrm{d}y\,\frac{a}{1-ay}\left[-\ln{\left(1-by\right)}\right]\\
&=\int_{0}^{1}\mathrm{d}y\,\frac{a}{1-ay}\int_{0}^{1}\mathrm{d}x\,\frac{by}{1-byx}\\
&=\int_{0}^{1}\mathrm{d}y\int_{0}^{1}\mathrm{d}x\,\frac{aby}{\left(1-ay\right)\left(1-bxy\right)}\\
&=\int_{0}^{1}\mathrm{d}x\int_{0}^{1}\mathrm{d}y\,\frac{aby}{\left(1-ay\right)\left(1-bxy\right)}\\
&=\int_{0}^{1}\mathrm{d}x\int_{0}^{1}\mathrm{d}y\,\frac{ab}{\left(a-bx\right)}\left[\frac{1}{\left(1-ay\right)}-\frac{1}{\left(1-bxy\right)}\right]\\
&=\int_{0}^{1}\mathrm{d}x\,\frac{ab}{\left(a-bx\right)}\left[\int_{0}^{1}\mathrm{d}y\,\frac{1}{\left(1-ay\right)}-\int_{0}^{1}\mathrm{d}y\,\frac{1}{\left(1-bxy\right)}\right]\\
&=\int_{0}^{1}\mathrm{d}x\,\frac{ab}{\left(a-bx\right)}\left[-\frac{1}{a}\int_{0}^{1}\mathrm{d}y\,\frac{(-a)}{\left(1-ay\right)}+\frac{1}{bx}\int_{0}^{1}\mathrm{d}y\,\frac{(-bx)}{\left(1-bxy\right)}\right]\\
&=\int_{0}^{1}\mathrm{d}x\,\frac{ab}{\left(a-bx\right)}\left[-\frac{1}{a}\ln{\left(1-a\right)}+\frac{1}{bx}\ln{\left(1-bx\right)}\right]\\
&=\int_{0}^{1}\mathrm{d}x\,\left[\frac{a\ln{\left(1-bx\right)}}{x\left(a-bx\right)}-\frac{b\ln{\left(1-a\right)}}{\left(a-bx\right)}\right]\\
&=\int_{0}^{1}\mathrm{d}x\,\left[\frac{\ln{\left(1-bx\right)}}{x}+\frac{b\ln{\left(1-bx\right)}}{\left(a-bx\right)}-\frac{b\ln{\left(1-a\right)}}{\left(a-bx\right)}\right]\\
&=\int_{0}^{1}\mathrm{d}x\,\left[\frac{\ln{\left(1-bx\right)}}{x}+\frac{b\ln{\left(\frac{1-bx}{1-a}\right)}}{\left(a-bx\right)}\right]\\
&=-\int_{0}^{1}\mathrm{d}x\,\frac{(-1)\ln{\left(1-bx\right)}}{x}+\int_{0}^{1}\mathrm{d}x\,\frac{b\ln{\left(\frac{1-bx}{1-a}\right)}}{\left(a-bx\right)}\\
&=-\operatorname{Li}_{2}{\left(b\right)}+\int_{0}^{1}\mathrm{d}x\,\frac{b\ln{\left(\frac{1-bx}{1-a}\right)}}{\left(a-bx\right)}\\
&=-\operatorname{Li}_{2}{\left(b\right)}+\int_{\frac{1}{1-a}}^{\frac{1-b}{1-a}}\mathrm{d}t\,\frac{\ln{\left(t\right)}}{\left(1-t\right)};~~~\small{\left[x=\frac{1-(1-a)t}{b}\right]}\\
&=-\operatorname{Li}_{2}{\left(b\right)}+\int_{\frac{a}{a-1}}^{\frac{a-b}{a-1}}\mathrm{d}u\,\frac{(-1)\ln{\left(1-u\right)}}{u};~~~\small{\left[t=1-u\right]}\\
&=\operatorname{Li}_{2}{\left(\frac{a-b}{a-1}\right)}-\operatorname{Li}_{2}{\left(\frac{a}{a-1}\right)}-\operatorname{Li}_{2}{\left(b\right)}.\blacksquare\\
\end{align}$$

A: We have:
$$ I'(c) = \int_{0}^{1}\frac{x\,dx}{(1+x)(c x-1)}=\frac{\log 2}{1+c}+\frac{\log(1-c)}{c+c^2}$$
and since $I(0)=0$, it follows that:
$$ I(c) = \log(2) \log(1+c)+\int_{0}^{c}\frac{\log(1-x)}{x}\,dx-\int_{0}^{c}\frac{\log(1-x)}{1+x}\,dx$$
so:

$$ I(c) = \text{Li}_2\left(\frac{1+c}{2}\right)-\text{Li}_2(c)+\text{Li}_{2}\left(\frac{1}{2}\right)$$
  where $\text{Li}_2\left(\frac{1}{2}\right)=\frac{\pi^2}{12}-\frac{1}{2}\log^2 2$.

That is straightforward to check through differentiation, too.
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,}
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 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
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 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
\mrm{I}\pars{c} & =
\int_{0}^{1}{\ln\pars{1 - cx} \over 1 + x}\,\dd x
\,\,\,\stackrel{1 - cx\ \mapsto\ x}{=}\,\,\,
-\int_{1}^{1 - c}{\ln\pars{x} \over 1 + c - x}\,\dd x
\\[5mm] & \stackrel{x/\pars{1 + c}\ \mapsto\ x}{=}\,\,\,
-\int_{1/\pars{1 + c}}^{\pars{1 - c}/\pars{1 + c}}
{\ln\pars{\bracks{c + 1}x} \over 1 - x}\,\dd x
\\[5mm] & = 
\left.\vphantom{\LARGE A}\ln\pars{1 - x}\ln\pars{\bracks{c + 1}x}
\,\right\vert_{\ 1/\pars{1 + c}}^{\ \pars{1 - c}/\pars{1 + c}} -
\int_{1/\pars{1 + c}}^{\pars{1 - c}/\pars{1 + c}}{\ln\pars{1 - x} \over x}
\,\dd x
\\[5mm] & =
\ln\pars{1 - {1 - c \over 1 + c}}\ln\pars{1 - c} +
\int_{1/\pars{1 + c}}^{\pars{1 - c}/\pars{1 + c}}\mrm{Li}_{2}'\pars{x}\,\dd x
\\[5mm] & =\ \bbox[#ffe,15px,border:1px dotted navy]{\ds{%
\ln\pars{2c \over 1 + c}\ln\pars{1 - c} +
\mrm{Li}_{2}\pars{1 - c \over 1 + c} - \mrm{Li}_{2}\pars{1\over 1 + c}}}
\end{align}
