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

Question

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!

Answer 1

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?