<s>[INST] Weird Integration problem: $\int_{-2}^{2} \frac{x^2}{1+5^x}dx$ $\int_{-2}^{2} \frac{x^2}{1+5^x}dx$
I am stuck at the first step and have tried replacing $5^x$ with $e^{\ln(5^x)}$ but nothing simplifies out in the end. 
Any hints how I should proceed?
 [/INST] $$I=\int_{-2}^{2} \frac{x^2}{1+5^x}dx$$
Let $y=-x$, then:
$$I=\int_{-2}^{2} \frac{y^2}{1+5^{-y}}dy=\int_{-2}^{2} \frac{y^25^y}{1+5^y}dy=\int_{-2}^{2} \frac{x^25^x}{1+5^x}dx$$
So:
$$I+I=\int_{-2}^{2} \frac{x^2}{1+5^x}dx+\int_{-2}^{2} \frac{x^25^x}{1+5^x}dx=\int_{-2}^2x^2\,dx=2\int_0^2x^2\,dx=2\left(\frac{2^3}{3}\right)=\frac{16}{3}$$
So, the integral is equal to $\dfrac{8}{3}$.
Edit: A clarification on the changing of limits under $y=-x$:
$$\int_{x=-2}^{x=2}f(x)\,dx=\int_{-y=-2}^{-y=2}f(-y)(-\,dy)=-\int_{y=2}^{y=-2}f(-y)\,dy=\int_{y=-2}^{y=2}f(-y)\,dy$$
$$=\int_{-2}^{2}f(-y)\,dy$$
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