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.