How to compute the integral $ I\left(c\right)=\int_{0}^{1}{\frac{\ln(1-cx)}{1+x}dx} $
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.
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}$$