TITLE: Integrating $\int^2_{-2}\frac{x^2}{1+5^x}$
QUESTION [5 upvotes]: $$\int^2_{-2}\frac{x^2}{1+5^x}$$
How do I start to integrate this?
I know the basics and tried substituting $5^x$ by $u$ where by changing the base of logarithm I get $\frac{\ln(u)}{\ln 5}=x$, but I got stuck.
Any hints would suffice preferably in the original question and not after my substitution.
(And also using the basic definite integrals property.)
Now I know only basic integration, that is restricted to high school, so would prefer answer in terms of that level.

REPLY [9 votes]: $$\tag1I=\int_{-2}^{2}\frac{x^2}{1+5^x}dx$$
Note that $$\int_a^bf(x)dx=\int_a^bf(a+b-x)dx$$
Thus,
$$\tag2I=\int_{-2}^{2}\frac{(-2+2-x)^2}{1+5^{-2+2-x}}dx=\int_{-2}^{2}\frac{x^2}{1+5^{-x}}dx=\int_{-2}^{2}\frac{5^xx^2}{1+5^{x}}dx$$
Add $(1)$ and $(2)$.