Let $[-a, a] \subset \mathbb{R}$ be an interval. Let $f: [-a, a] \rightarrow \mathbb{R}$ be a Riemann-integrable function such that $f(x) = f(-x)$ for all $x \in [-a, a]$. Let $b \in \mathbb{R}_+$.
Question: Is there a "good" way to simplify the integral $\int_{-a}^a \frac{f(x) + 1}{b^x+1} \,\mathrm dx$?
Yes! In fact,
$$\displaystyle\int_{-a}^a \frac{f(x)+1}{b^x+1} \,\mathrm dx= a+\int_0^a f(x) \,\mathrm{d}x.$$
Proof. We have \begin{equation}\label{*}\tag{*} \int_{-a}^a \frac{f(x)+1}{b^x+1} \,\mathrm dx \overset{\text{substituting } x \rightarrow -x}{=} (-1)^2 \cdot \int_{-a}^a \frac{f(x)+1}{b^{-x}+1} \,\mathrm{d}x = \int_{-a}^a \frac{b^x \cdot (f(x) + 1)}{b^x + 1} \,\mathrm{d}x \end{equation}
Thus \begin{split} \displaystyle 2 \cdot \int_{-a}^a \frac{f(x)+1}{b^x+1} \,\mathrm{d}x &\overset{\eqref{*}}{=} \int_{-a}^a \frac{f(x)+1}{b^x+1} \,\mathrm{d}x + \int_{-a}^a \frac{b^x \cdot (f(x)+1)}{b^x+1} \,\mathrm{d}x \\ &= \int_{-a}^a \frac{(b^x+1) \cdot (f(x)+1)}{b^x+1} \,\mathrm{d}x \\ &= \int_{-a}^a f(x)+1 \,\mathrm{d}x = 2 a + \int_{-a}^a f(x) \,\mathrm{d}x \end{split}
This achieves a proof. $\square$