Showing $\int_{0}^1 \frac{\log x}{x+1} dx=-\int_{0}^1 \frac{\log(x+1)}{x} dx$

We can show that $$\int_{0}^1 \frac{\log x}{x+1} dx=-\int_{0}^1 \frac{\log(x+1)}{x} dx \tag{*}$$ by evaluating these two integral equal to $$\sum_{k=1}^{\infty} \frac{(-1)^{k+1} }{k^2}-\frac{\pi^2}{12}$$ by expanding $(1+x)^{-1}$ and $\log (1+x)$ in Maclaurin series. The question is: How to show $(*)$ without series expansions.