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.
The integral of $f(x)$ from $-2$ to $2$ is always the same as the integral of $f(-x)$ from $-2$ to $2$, so the integral is the same as $$\int_{-2}^2 {1 + x^2 \over 1 + 2^{-x}}\,dx$$ Multiplying the numerator and denominator by $2^x$ gives $$= \int_{-2}^2 {2^x (1 + x^2) \over 1 + 2^x}\,dx$$ Adding this to the orginal expression, twice the integral is equal to $$\int_{-2}^2 {1 + x^2 \over 1 + 2^x}\,dx + \int_{-2}^2 {2^x (1 + x^2) \over 1 + 2^x}\,dx$$ $$= \int_{-2}^2 {(1 + 2^x)(1 + x^2) \over 1 + 2^x}\,dx$$ $$= \int_{-2}^2 ({1 + x^2})\,dx$$ $$= {28\over 3}$$ Hence the original integral is half that, or ${\displaystyle {14 \over 3}}$.
Let $$I=\int_{-2}^{2} \frac{1+x^2}{1+2^x}dx$$ we split it into two integrals $$I_1=\int_{-2}^{0} \frac{1+x^2}{1+2^x}dx$$ and $$I_2=\int_{0}^{2} \frac{1+x^2}{1+2^x}dx$$ On $I_1$ use the substitution $u=-x$. Proceed by setting $I=I_1 +I_2$ and applying the hint given by Ron Gordon.
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 $\displaystyle{\frac{14}{3}}$.
