parametric spiral arc length I am attempting to calculate the arc length of a spiral. Given an Archamedean spiral of the parametric form:
$x(t) = \sqrt{t}\cos\left(\omega\sqrt{t} \right)$ and $y(t) = \sqrt{t}\sin\left(\omega\sqrt{t} \right)$, the arc length $L$ is calculated as
\begin{align}\label{parametricArcLength}
 L & = \int_a^b \sqrt{1 + \left(\frac{\frac{dy}{dt}}{\frac{dx}{dt}} \right)^2}\cdot \frac{dx}{dt}dt \\
 & = \int_a^b \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \ \ dt 
\end{align}
with 
\begin{align}
 \frac{dx}{dt} & = \frac{1}{2\sqrt{t{}}} \cdot \cos (\omega \sqrt{t}) - \sqrt{t} \cdot \frac{\omega}{2\sqrt{t}} \sin (\omega \sqrt{t}) \\
 & = \frac{\cos (\omega \sqrt{t})}{2\sqrt{t}} - \frac{\omega \sin (\omega \sqrt{t})}{2} \nonumber
\end{align}
\begin{align} \label{dy_dt}
\frac{dy}{dt} & = \frac{1}{2\sqrt{t}} \cdot \sin (\omega \sqrt{t}) + \sqrt{t} \cdot \frac{\omega}{2\sqrt{t}} \cos (\omega \sqrt{t}) \\
 & = \frac{\sin (\omega \sqrt{t})}{2 \sqrt{t}} + \frac{\omega \cos (\omega \sqrt{t})}{2} \nonumber
\end{align}
yielding
\begin{split}
L = &\int_{t_a}^{t_b} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \ \ dt \\
   = & \int_{t_a}^{t_b} \left[\frac{1}{4}\left(\frac{\cos^2 \left(\omega \sqrt{t}\right)}{t} - \frac{2\omega\cos\left(\omega\sqrt{t} \right) \sin \left(\omega \sqrt{t}\right)}{\sqrt{t}} + \omega^2 \sin^2\left(\omega \sqrt{t}\right)\right) \right.\\
 & \left. + \frac{1}{4}\left( \frac{\sin^2\left( \omega\sqrt{t} \right)}{t} + \frac{2\omega \sin \left(\omega \sqrt{t}\right) \cos\left(\omega \sqrt{t} \right)}{\sqrt{t}} + \omega^2 \cos\left(\omega \sqrt{t} \right)  \right) \right]^{1/2}dt\\
 = & \int_{t_a}^{t_b} \left[\frac{1}{4}\left( \frac{\sin^2\left(\omega\sqrt{t}\right) + \cos^2\left(\omega\sqrt{t}\right)}{t}   + \omega^2\sin^2\left(\omega\sqrt{t}\right) + \omega^2\cos^2\left(\omega\sqrt{t}\right)\right)\right]^{1/2}dt\\
 = & \int_{t_a}^{t_b} \left[ \frac{1}{4} \left( \frac{1}{t} + \omega^2 \right)\right]^{1/2}dt\\
 = &  \frac{1}{2}\int_{t_a}^{t_b}\left[\frac{1}{t} + \omega^2 \right]^{1/2}dt
\end{split}
Is the above integral the correct solution? If so, what method should be used to solve it?

Parametric curves are just screaming out to be solved in the complex plane. Consider that
$$z=x+iy=\sqrt{t}~e^{i\omega\sqrt{t}}$$
The arc length is given by
$$s=\int |\dot z|dt$$
Thus,
$$
\dot z=\left(\frac{1}{2\sqrt{t}}+\frac{i\omega}{2} \right)e^{i\omega\sqrt{t}}\\
|\dot z|=\frac{1}{2}\sqrt{\frac{1}{t}+\omega^2}
$$
Then
$$
\begin{align}
s=\int |\dot z|dt
&=\frac{1}{2}\int \sqrt{\frac{1}{t}+\omega^2}~dt\quad \text{(same as in the OP)}\\
&=\frac{1}{\omega}\int \sqrt{1+u^2}~du,\quad u=\omega\sqrt{t}\\
&=\frac{1}{2\omega}\left[u\sqrt{1+u^2}+\sinh^{-1}u \right]\\
&=\frac{1}{2\omega}\left[\omega\sqrt{t}\sqrt{1+\omega^2t}+\sinh^{-1}(\omega\sqrt{t}) \right]\\
\end{align}
$$
Notice, in particular, how this method circumvents all of the trigonometry in order to arrive at the same expression for the arc length integral as in the OP.
I have verified this result numerically for randomly selected values of $\omega$.
