Evaluate the integral $\int_{-2}^{2} \frac{1+x^2}{1+2^x}dx$ My friend asked me ot evaluate the integral:
$$\int_{-2}^{2} \frac{1+x^2}{1+2^x}dx$$
And he gave me the hint: substitute $u = -x$. And so I did that, but I can't seem to get any farther than that. Could someone please provide some hints and help as to how to evaluate this challenging integral?
EDIT: Another hint he gave me was the split the integral into 2 integrals, one from $-2$ to $0$ and the other from $0$ to $2$, and again, I have tried this and I get stuck.

Substituting yields $$\int_{2}^{-2}-\frac{1+x^2}{1+2^{-x}},$$ which we can add to the original integral to get $$\int_{-2}^{2} \frac{1+x^2}{1+2^x}dx + \int_{2}^{-2}-\frac{1+x^2}{1+2^{-x}} = \int_{-2}^{2}\frac{1+x^2}{1+2^x} + \frac{1+x^2}{1+2^{-x}} = \int_{-2}^{2}1+x^2 = \frac{28}{3},$$ so our original integral is half that, which is $\displaystyle{\frac{14}{3}}$.
