Area of a surface of revolution of $y = \sqrt{4x+1}$ $y = \sqrt{4x+1}$ for $1 \leq x \leq 5$
I really have no idea what to do with this problem, I attempted something earlier which I will not type up because it took me two pages. 
$$y = \sqrt{4x+1}$$
$$\int 2 \pi \sqrt{4x+1} \sqrt{1 + \frac{4}{1+4x}}dx$$
$$2 \pi \int \sqrt{4x+1} \sqrt{1 + \frac{4}{1+4x}}dx$$
Nothing really seems obvious at this point, I attempted a u substitution of $u = 1+4x$ but it does not help simplify this problem really. 
$$ \pi /2 \int \sqrt{u} \sqrt{1 + \frac{4}{u}}du$$
I thought about making a wonky trig substitution but it didn't seem to help and was overly complicated. 

From the last step:: 
$$ \frac{\pi}{2} \int \sqrt u \frac{\sqrt{4 +u}}{\sqrt u} du  = \frac \pi 2 \int \sqrt{4 + u} du $$
substituting $4 + u = p \implies  du = dp \;\;$, we get
$$ = \frac \pi 2 \int \sqrt p dp = \frac \pi 2 \frac{p^{3/2}}{3/2}  = \frac \pi 3 (4+u)^{3/2} $$
