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?

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