An easy method to solve this definite integral? I have this definite integral
$$ \int_{-3}^3 \frac{1+x^2}{1+2^x} \, dx$$
But I'm not sure how to calculate this integral in an easy and direct way.
Using calculators like Wolfram Alpha I found the answer to be 12 but I needed some hints on how to solve this further using pen and paper.
 A: Given an even function $e(x)$, an odd function $o(x)$ and parameters $a,b>0$, we will show that $$\int_{-a}^a\frac{e(x)}{1+b^{o(x)}}\,\mathrm dx=\int_0^ae(x)\,\mathrm dx,$$ which gives you a very quick method of evaluating your integral.

Set $f(x)=\frac{e(x)}{1+b^{o(x)}}$ as well as $E(x)=\frac12(f(x)+f(-x))$ and $O(x)=\frac12(f(x)-f(-x))$. This is the even-odd-decomposition of $f$, meaning $f=E+O$, $E$ is even and $O$ is odd. Thus we get \begin{align*}\int_{-a}^af(x)\,\mathrm dx&=\int_{-a}^aE(x)\,\mathrm dx=\int_{0}^a(f(x)+f(-x))\,\mathrm dx\\&=\int_0^ae(x)\left(\frac{1}{1+b^{o(x)}}+\frac1{1+b^{-o(x)}}\right)\,\mathrm dx=\int_0^ae(x)\,\mathrm dx.&\blacksquare\end{align*}

Now just notice that your integral is of the given form with $a=3,b=2,e(x)=1+x^2,o(x)=x$, and evaluate $\int_0^3(1+x^2)\,\mathrm dx$.
A: The substitution $y=-x$ yields
$$
\int_{-3}^3 \frac{1+x^2}{1+2^x} dx = \int_{-3}^3 \frac{1+y^2}{1+2^{-y}} dy.
$$
Thus
$$
2\int_{-3}^3 \frac{1+x^2}{1+2^x} dx = \int_{-3}^3 (1+x^2)\left(\frac{1}{1+2^x} + \frac{1}{1+2^{-x}}\right) dx
$$
$$
= \int_{-3}^3 (1+x^2) dx = 2\int_{0}^3 (1+x^2) dx= 2(3+27/3)=24.
$$
Thus indeed
$$
\int_{-3}^3 \frac{1+x^2}{1+2^x} dx =12
$$
A: Use the fact that $\int_a^b f(x)dx $ is same as $\int_a^b f(a+b-x)dx $
Define, $$ I = \int_{-3}^3 {{1+x^2}\over{1+2^x}}dx $$
Now, use the fact stated above and add both the integrals to get
$$2I = \int_{-3}^3 {1+x^2} dx $$
Evaluating which, should be pretty straightforward.
A: Note that $\frac1{1+2^x} =\frac12-\frac12\tanh\frac {x\ln2}2$, where only the even term $\frac12$ contributes to the integral, i.e.
$$ \int_{-3}^3 \frac{1+x^2}{1+2^x} \, dx=\frac12 \int_{-3}^3 (1+x^2)\, dx = 12
$$
A: $$I=\int_{-3}^3\frac{1+x^2}{1+2^x}dx=\int_{-3}^3\frac{1+x^2}{1+2^{-x}}dx$$
$$2I=\int_{-3}^3\left(\frac1{1+2^x}+\frac{1}{1+2^{-x}}\right)(1+x^2)dx$$
now use the fact that:
$$\frac1{1+2^x}+\frac{1}{1+2^{-x}}=\frac1{1+2^x}+\frac{2^x}{1+2^x}=\frac{1+2^x}{1+2^x}=1$$
which gives you:
$$I=\frac12\int_{-3}^3(1+x^2)dx=\int_0^3(1+x^2)dx$$
