I'm lost as to how to go about solving this. u = 4x and du = 4 dx but beyond that I'm stumped. That constant in the denominator confuses me.
Now use a trigonometric substitution $\displaystyle u = 8\tan{\theta}$. Make note that $\displaystyle du = 8\sec^2{\theta}\,d\theta$.
Also note that $\displaystyle \theta = \arctan{\frac{u}{8}}$.
$\displaystyle \int{\frac{1}{u^2 + 64}\,du} = \int{\frac{1}{(8\tan{\theta})^2 + 64}\,8\sec^2{\theta}\,d\theta}$
$\displaystyle = \int{\frac{8\sec^2{\theta}}{64\tan^2{\theta} + 64}\,d\theta}$
$\displaystyle = \int{\frac{8\sec^2{\theta}}{64(\tan^2{\theta} + 1)}\,d\theta}$
$\displaystyle = \int{\frac{\sec^2{\theta}}{8\sec^2{\theta}}\,d\the ta}$
$\displaystyle = \int{\frac{1}{8}\,d\theta}$
$\displaystyle = \frac{1}{8}\theta + C$
$\displaystyle = \frac{1}{8}\arctan{\frac{u}{8}} + C$
$\displaystyle = \frac{1}{8}\arctan{\frac{e^{2x}}{8}} + C$.
Because if you have an integral of the form
$\displaystyle \int{\frac{1}{a^2 + x^2}\,dx}$
you make the substitution $\displaystyle x = a\tan{\theta}$.
This is so that you can factor out the $\displaystyle a^2$ and then make use of the identity $\displaystyle 1 + \tan^2{\theta} = \sec^2{\theta}$.
And since the derivative of $\displaystyle \tan{\theta}$ is $\displaystyle \sec^2{\theta}$, this also means that the $\displaystyle \sec^2{\theta}$ will be eliminated (as you will end up with them on the top and bottom).