Show that $\int_0^1 \frac{\ln(1+x)}x\mathrm dx=-\frac12\int_0^1 \frac{\ln x}{1-x}\mathrm dx$ without actually evaluating both integrals While doing some research on the 'alternating Basel Problem' I have come across this related post which states the equality

$$\int_0^1 \frac{\ln(1+x)}x\mathrm dx=-\frac12\int_0^1 \frac{\ln x}{1-x}\mathrm dx\tag1$$

Using the Dilogarithm one can show that 'alternating Basler Problem' is a direct consequence of this equation and yields to
$$\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n^2}=\frac{\pi^2}{12}$$
Therefore I have no doubts to trust the author of the the cited post. However, I tried to verify the equality by myself and failed. For this purpose I enforced the substitution $x\mapsto1+x$ within the integral on the right
$$\begin{align}
-\frac12\int_0^1 \frac{\ln x}{1-x}\mathrm dx=-\frac12\int_{(0-1)}^{(1-1)} \frac{\ln(1+x)}{1-(1+x)}\mathrm dx=-\frac12\int_{-1}^{0} \frac{\ln(1+x)}x\mathrm dx
\end{align}$$
But from hereon I am not sure how to proceed. Clearly now I have to show that
$$\begin{align}
-\frac12\int_{-1}^0\frac{\ln(1+x)}x\mathrm dx&=\int_0^1 \frac{\ln(1+x)}x\mathrm dx\\
\frac12\int_0^1\frac{\ln(1-x)}x\mathrm dx&=\int_0^1 \frac{\ln(1+x)}x\mathrm dx\\
0&=\int_0^1 \frac1x\left(\ln(1+x)-\frac12\ln(1-x)\right)\mathrm dx
\end{align}$$
It seems like I have made a mistake somewhere inbetween since WolframAlpha does not agree with my reasoning. Additionally I have no idea how to proceed. To be honest I am quite confused right now.

First of all where exactly did I went wrong? Furthermore could someone provide a complete proof for the given equality? Please tell me when this question has been asked before.

Thanks in advance!
 A: \begin{align}
\int_0^1\frac{\ln(1+x)}{x}dx&=\int_0^1\frac{\ln\left(\frac{1-x^2}{1-x}\right)}{x}dx\\
&=\underbrace{\int_0^1\frac{\ln(1-x^2)}{x}dx}_{1-x^2-\to x}-\underbrace{\int_0^1\frac{\ln(1-x)}{x}dx}_{1-x\to x}\\
&=\frac12\int_0^1\frac{\ln(x)}{1-x}dx-\int_0^1\frac{\ln(x)}{1-x}dx\\
&=-\frac12\int_0^1\frac{\ln x}{1-x}dx.
\end{align}
A: HINT: 
Note that we have
$$\begin{align}
\frac12\int_0^1 \frac{\log(x)}{1-x}\,dx&\overbrace{=}^{x\mapsto x^2}\int_0^1 \frac{x\log(x^2)}{1-x^2}\,dx\\\\
&=\int_0^1 \log(x)\left(\frac{1}{1-x}-\frac{1}{1+x}\right)\,dx
\end{align}$$
Can you finish now?
A: Show
$\int_0^1 \frac{\ln(1+x)}x\mathrm dx
=-\frac12\int_0^1 \frac{\ln x}{1-x}\mathrm dx
$
Playing around with series expansions.
$\begin{array}\\
I_1
&=\int_0^1 \frac{\ln(1+x)}x dx\\
&=\int_0^1 \sum_{n=0}^{\infty} \dfrac{(-1)^{n}x^n}{n+1}dx\\
&=\sum_{n=0}^{\infty} \dfrac{(-1)^{n}x^{n+1}}{(n+1)^2}|_0^1\\
&=\sum_{n=0}^{\infty} \dfrac{(-1)^{n}}{(n+1)^2}\\
I_2
&=\int_0^1 \frac{\ln x}{1-x}dx\\
&=\int_0^1 \frac{\ln (1-x)}{1-(1-x)}dx\\
&=\int_0^1 \frac{\ln (1-x)}{x}dx\\
&=-\int_0^1 \sum_{n=0}^{\infty}\dfrac{x^n}{n+1}dx\\
&=-\sum_{n=0}^{\infty}\int_0^1 \dfrac{x^n}{n+1}dx\\
&=-\sum_{n=0}^{\infty}\dfrac1{(n+1)^2}\\
2I_1+I_2
&=\sum_{n=0}^{\infty} \dfrac{2(-1)^{n}-1}{(n+1)^2}\\
&=\sum_{n=0}^{\infty} \dfrac{2(-1)^{2n}-1}{(2n+1)^2}
+\sum_{n=0}^{\infty} \dfrac{2(-1)^{2n+1}-1}{(2n+2)^2}\\
&=\sum_{n=0}^{\infty} \dfrac{1}{(2n+1)^2}
+\sum_{n=0}^{\infty} \dfrac{-3}{(2n+2)^2}\\
&=\sum_{n=0}^{\infty} \dfrac{1}{(2n+1)^2}
+\sum_{n=0}^{\infty} \dfrac{1}{(2n+2)^2}
+\sum_{n=0}^{\infty} \dfrac{-4}{(2n+2)^2}\\
&=\sum_{n=0}^{\infty} \dfrac{1}{(n+1)^2}
-\sum_{n=0}^{\infty} \dfrac{4}{4(n+1)^2}\\
&=\sum_{n=0}^{\infty} \dfrac{1}{(n+1)^2}
-\sum_{n=0}^{\infty} \dfrac{1}{(n+1)^2}\\
&=0\\
\end{array}
$
And it works!
