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.

All you need is that $$\int a^x dx = \int \exp(x \log a) dx = \frac{\exp(x \log a)}{\log a} +C= \frac{a^x}{\log(a)}+C$$
$$\begin{align*}
\int \frac{1+2^{2x}}{2^{x}} dx &= \int \left(\left(\frac12 \right)^x + 2^x \right)dx\\
 &= \left( \frac{\left(\frac12 \right)^x}{\log(1/2)} + \frac{2^x}{\log(2)} \right) + C\\
 &= \left( -\frac{\left(\frac12 \right)^x}{\log(2)} + \frac{2^x}{\log(2)} \right) + C\\
 &= \frac{\left(2^x - \frac1{2^x} \right)}{\log(2)} + C\end{align*}$$
