Evaluate the integral $\int_{-1}^1 x \ln \frac{x}{1-e^{-a x}} dx$

Symmetry is the key. Let $f_a(x)=x \log\frac{x}{1-e^{-a x}}$. We may notice that

$$ f_a(-x) = -x \log \frac{x}{e^{ax}-1} = -x\log\frac{x}{1-e^{-ax}}-x\log\frac{1}{e^{ax}}=-f_a(x)+ax^2 \tag{1}$$ hence: $$ \int_{-1}^{1}f_a(x)\,dx = \int_{0}^{1}\left(f_a(x)+f_a(-x)\right)\,dx=\int_{0}^{1}ax^2\,dx =\color{red}{\frac{a}{3}}.\tag{2}$$