Evaluate the integral $\int_{-1}^1 x \ln \frac{x}{1-e^{-a x}} dx$ 
How would you evaluate this definite integral with a combination of logarithmic and exponential functions:
$$I(a)=\int_{-1}^1 x \ln \frac{x}{1-e^{-a x}} dx, \qquad a \geq 0$$

Mathematica didin't solve it for the general case.
My solution is hidden below (spoiler tag doesn't work properly for multiple lines for some reason).
Substitute $x=-t$:
$$I(a)=\int_{-1}^1 t \ln \frac{e^{a t}-1}{t} dt$$

 $$2 I(a)=\int_{-1}^1 x \ln \frac{e^{a x}-1}{1-e^{-a x}} dx=a \int_{-1}^1 x^2 dx+\int_{-1}^1 x \ln 1 dx=\frac{2a}{3}$$

The answer is:
$$I(a)=\frac{a}{3}$$
That's not how I actually found this solution though.
But how would you, at a first glance, attempt to solve it?
 A: 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:
$$ \int_{-1}^{1}f_a(x)\,dx = \int_{0}^{1}\left(f_a(x)+f_a(-x)\right)\,dx=\int_{0}^{1}ax^2\,dx =\color{red}{\frac{a}{3}}.\tag{2}$$
A: A slight generalisation $(q>0,n\in\mathbb{N})$. Let us denote
$$\
I_{q,n}(a)=\int_{-q}^qdxx^{2n-1}\log\left(\frac{x}{1-e^{-ax}}\right)
$$
Now differentiate w.r.t $a$. We get
$$
I_{q,n}'(a)=\int_{-q}^{q}dx\frac{x^{2n}}{1-e^{ax}}=\int_{0}^{q}dx\frac{x^{2n}}{1-e^{ax}}+\int_{-q}^{0}dx\frac{x^{2n}}{1-e^{ax}}=\\
=\int_{0}^{q}dx\frac{x^{2n}}{1-e^{ax}}+\int_{0}^{q}dx\frac{x^{2n}}{1-e^{-ax}}=\\
\int_{0}^{q}dx\frac{x^{2n}}{1-e^{ax}}-\int_{0}^{q}dx\frac{x^{2n}e^{ax}}{1-e^{ax}}=\\
\\\int_{0}^{q}dxx^{2n}=\frac{q^{2n+1}}{2n+1}
$$
so 
$$
I_{q,n}(a)=\frac{q^{2n+1}a}{2n+1}+C
$$
but by symmetry $I(0_+)=0$ and we get

$$
I_{q,n}(a)=\frac{q^{2n+1}a}{2n+1}
$$

and the actual question is answered by setting $q=1,n=1$

$$
I(a)=I_{1,1}(a)=\frac{a}{3}
$$

