Finding the integral of $(1+2^{2x})/2^x$ 
Evaluate the integral:
  $$\int\frac{1+2^{2x}}{2^x}\,dx = \int \frac{ 1 + (2^x)^2}{2^x}\,dx$$

Let $u = 2^x$.  Then $du = 2^x\ln2\,dx$, which yields $\frac{du}{2^x\ln2} = dx$ so
$$  \int \frac{ 1 + (2^x)^2}{2^x}\,dx =  \int \frac{1+u^2}{u}du =
\left( x+ \frac{u^3}{3} \right)\ln u+C$$
$$=\left(x+ \frac{(2^x)^3}{3} \right)\ln 2^x +C$$
I'm not sure if I integrated this correctly. Any help would be appreciated.

Your change of variable is fine; your substitution is not quite right and your integral is not quite right. If $u=2^x$, then $du = 2^x\ln(2)\,dx = u\ln(2)\,dx$, so $dx = \frac{du}{\ln(2)u}$. So we have:
$$\begin{align*}
\int \frac{1+2^{2x}}{2^x}\,dx &= \int\frac{1 + (2^x)^2}{2^x}\,dx\\
&= \int\left(\frac{1 + u^2}{u}\right)\frac{du}{\ln(2)u}\\
&= \frac{1}{\ln(2)}\int\frac{1+u^2}{u^2}\,du\\
&=\frac{1}{\ln(2)}\int\left(u^{-2} + 1\right)\,du\\
&=\frac{1}{\ln(2)}\left(-u^{-1} + u\right) + C\\
&= \frac{1}{\ln(2)}\left(2^x - \frac{1}{2^x}\right) + C
\end{align*}$$
