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showing $\int\limits_{-a}^a \frac{f(x)}{1+e^{x}} \mathrm dx = \int\limits_0^a f(x) \mathrm dx$ when $f$ is even

I have a question:

Suppose $f$ is continuous and even on $[-a,a]$, $a>0$ then prove that $$\int\limits_{-a}^a \frac{f(x)}{1+e^{x}} \mathrm dx = \int\limits_0^a f(x) \mathrm dx$$

How can I do this? Don't know how to start.

James
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