$\int_{-b}^{b} \frac{f(x)}{a^x+1} dx$ where $f:[-b,b]\rightarrow\mathbb{R}$ even function and $a,b>0$

Question

$$\int_{-b}^{b} \frac{f(x)}{a^x+1} dx$$ where $$f:[-b,b]\rightarrow\mathbb{R}$$ f is an even function and $a,b>0$. The answer should be : $$\int_{0}^{b} f(x)dx$$ Solve this without any high math level methods please because I'm only a highschooler, thanks.

Answer 1

Let $$I=\int_{-b}^b \dfrac{f(x)}{a^x+1}\,\mathrm dx\tag{1}$$ Using the fact that

$$\int_a^b f(x)\,\mathrm dx=\int_a^b f(a+b-x)\,\mathrm dx$$

we can write the integral as $$\begin{align}I&=\int_{-b}^b\dfrac{f(-x)}{a^{-x}+1}\,\mathrm dx\\&=\int_{-b}^b\dfrac{a^xf(x)}{a^x+1}\,\mathrm dx\qquad f(x)\text{ is even }\Longleftrightarrow f(-x)=f(x)\end{align}\tag{2}$$

Adding $(1)$ and $(2)$ $$\begin{align}2I&=\int_{-b}^b\dfrac{f(x)}{a^x+1}\,\mathrm dx+\int_{-b}^b\dfrac{a^xf(x)}{a^x+1}\,\mathrm dx\\&=\int_{-b}^b\dfrac{(a^x+1)f(x)}{a^x+1}\,\mathrm dx\\&=\int_{-b}^bf(x)\,\mathrm dx\\&=2\int_0^b f(x)\,\mathrm dx\qquad\text{Because }f(x)\text{ is even about }x=0\\I&=\int_0^b f(x)\,\mathrm dx\end{align}$$