Substitution

Relevant wiki: Integration Tricks

$\begin{aligned} I & = \int_{-2}^2 \frac {1+x^2}{1+2^x} dx & \small \color{#3D99F6} \text{Using identity }\int_a^b f(x) \ dx = \int_a^b f(a+b-x) \ dx \\ & = \frac 12 \int_{-2}^2 \frac {1+x^2}{1+2^x} + \frac {1+x^2}{1+2^{-x}} dx \\ & = \frac 12 \int_{-2}^2 \left(1+x^2\right) \left( \frac 1{1+2^x} + \frac {2^x}{2^x+1} \right) dx \\ & = \frac 12 \int_{-2}^2 \left(1+x^2\right) \ dx & \small \color{#3D99F6} \text{Note that } 1+x^2 \text{ is even.} \\ & = \int_0^2 \left(1+x^2\right) \ dx \\ & = x + \frac {x^3}3 \bigg|_0^2 = 2 + \frac 83 = \frac {14}3 \end{aligned}$

$\implies a + b = 14+3 = \boxed{17}$

Relevant wiki: Integration Tricks

We substitute the variable $x$ with $- x$ and add the resulting integral to the original integral to get

$\displaystyle 2I = \int_{- 2}^2 \frac {1 + x^2}{1 + 2^x} \ dx + \int_2^{- 2} - \frac {1 + x^2}{1 + 2^{- x}} \ dx$

$\displaystyle = \int_{- 2}^2 \frac {1 + x^2}{1 + 2^x} + \frac {1 + x^2}{1 + 2^{- x}} \ dx$

$\displaystyle = \int_{- 2}^2 \frac {1 + x^2}{1 + 2^x} + \frac {1 + x^2}{1 + 2^{- x}} \cdot \frac{2^x}{2^x} \ dx$

$\displaystyle = \int_{- 2}^2 \frac {(1 + x^2)2^x}{1 + 2^x} + \frac {1 + x^2}{1 + 2^x} \ dx$

$\displaystyle \int_{- 2}^2 \frac {(1 + x^2) \cdot (1 + 2^x)}{1 + 2^x} \ dx$

$\displaystyle \int_{- 2}^2 1 + x^2 \ dx = 4 + \frac {16}{3} = \frac {28}{3}$

Thus, $\displaystyle I = \large{\boxed{\frac {14}{3}}}$

Note that more generally, for even functions $f$ , $\displaystyle \int_{- a}^a \frac{f(x)}{1 + b^x} \ dx = \frac 12 \int_{- a}^a f(x) \ dx$ .