FGCU Math Competition Problem

Relevant wiki: Integration Tricks

$\begin{aligned} I & = \int_{-2}^2 \frac {1+x^2}{1+2^x}dx & \small \color{#3D99F6} \text{By identity }\int_a^b f(x) \ dx = \int_a^b f(a+b-x) \ dx \\ & = \frac 12 \int_{-2}^2 \left(\frac {1+x^2}{1+2^x} + \frac {1+x^2}{1+2^{-x}} \right) dx \\ & = \frac 12 \int_{-2}^2 \left(\frac {1+x^2}{1+2^x} + \frac {2^x(1+x^2)}{2^x+1} \right) dx \\ & = \frac 12 \int_{-2}^2 \left(1+x^2 \right) dx & \small \color{#3D99F6} \text{Note that the integrand is even.} \\ & = \int_0^2 \left(1+x^2 \right) dx \\ & = x + \frac {x^3}3 \ \bigg|_0^2 \\ & = 2 + \frac 83 \\ & \approx \boxed{4.67} \end{aligned}$