[Some context below, actual question at the end]
Wolfram MathWorld defines the paraboloid using the following parametrization ($u \geq 0$, $v \in [0,2\pi)$):
\begin{align} x&=\sqrt{u}\cos{v} \\ y&=\sqrt{u}\sin{v} \\ z&=u \end{align}
and the coefficients of the first fundamental form as
\begin{align} P&=\left(\frac{\partial x}{\partial u}\right)^2+\left(\frac{\partial y}{\partial u}\right)^2+\left(\frac{\partial z}{\partial u}\right)^2=1+\frac{1}{4u} \\[8pt] Q&=\frac{\partial x}{\partial u}\frac{\partial x}{\partial v}+\frac{\partial y}{\partial u}\frac{\partial y}{\partial v}+\frac{\partial z}{\partial u}\frac{\partial z}{\partial v}=0 \\[10pt] R&=\left(\frac{\partial x}{\partial v}\right)^2+\left(\frac{\partial y}{\partial v}\right)^2+\left(\frac{\partial z}{\partial v}\right)^2=u \end{align}
Next, setting up the Euler-Lagrange equation is pretty straightforward:
\begin{equation} \underbrace{\frac{\frac{\partial P}{\partial v}+2v\,'\frac{\partial Q}{\partial v}+v\,'^2\frac{\partial R}{\partial v}}{2\sqrt{P+2Qv\,'+Rv\,'^2}}}_{\displaystyle{=0}}-\frac{d}{du}\left(\frac{Q+Rv'}{\sqrt{P+2Qv\,'+Rv\,'^2}}\right)=0 \end{equation}
Plugging in $P$, $Q$ and $R$ from above, the equation will greatly simplify because of the vanishing term on the LHS, so we are left with
\begin{equation} \frac{d}{du}\left(\frac{u\,v'(u)}{\sqrt{1+\frac{1}{4u}+u\,v'(u)^2}}\right)=0 \end{equation}
Since the total derivative yields zero, the term in parentheses must be constant:
\begin{equation} \frac{u\,v'(u)}{\sqrt{1+\frac{1}{4u}+u\,v'(u)^2}}=C_1 \end{equation}
Rearrange:
\begin{align} u\,v'(u)&=C_1\sqrt{1+\frac{1}{4u}+u\,v'(u)^2} \\[8pt] u^2\,v'(u)^2&=C_1\left(1+\frac{1}{4u}+u\,v'(u)^2\right) \\[8pt] \end{align} (Leave $C_1$ unsquared since its value is arbitrary anyway)
Eventually I arrive at \begin{equation} v(u)=\pm\int \sqrt{\frac{C_1+\frac{C_1}{4u}}{u^2-C_1 u}}du \end{equation}
I was curious to see what Mathematica would come up with (solution for the positive branch):
Integrate[Sqrt[(C1 + C1/(4 u))/(u^2 - C1 u)], u,Assumptions -> u >= 0]
yields
\begin{gather*} v(u)=\frac{u\sqrt{-C_1+u}\sqrt{\frac{C_1(1+4u)}{u^2(-C_1+u)}}\Bigg(2\sqrt{C_1}\sqrt{C_1(1+4 C_1)}\sqrt{\frac{1+4u}{1+4 C_1}}\text{arcsinh}\left(\frac{2\sqrt{C_1}\sqrt{-C_1+u}}{\sqrt{C_1(1+4 C_1)}}\right) }{C_1+4 C_1\,u} \\[8pt] \frac{+\sqrt{C_1+4 C_1\,u}\,\arctan\left(\frac{\sqrt{-C_1+u}}{\sqrt{C_1+4 C_1\,u}}\right)\Bigg)}{C_1+4 C_1\,u} +C_2 \end{gather*}
Contrast this to the solution from MathWorld (eq. 11):
\begin{equation} u-c^2=u(1+4c^2)\sin^2\left(v-2c\ln\left(k\left(2\sqrt{u-c^2}+\sqrt{4u+1}\right)\right)\right) \end{equation}
where $c$ and $k$ are the integration constants.
Apart from some more or less obvious simplifications to the Mathematica output, I was wondering how Weinstock came up with his solution. Any hints most welcome!
Edit:
Using the identities suggested by Rafa, I was able to get closer to the solution given by Weinstock. First, the output from Mathematica can be further simplified to
$$ v=2\sqrt{C_1}\text{arcsinh}\left(\frac{2\sqrt{-C_1+u}}{\sqrt{1+4 C_1}}\right)+\arctan\left(\frac{\sqrt{-C_1+u}}{\sqrt{C_1+4C_1u}}\right)+C_2 $$
Now, use the trig identities
\begin{align*} \text{arcsinh}\left(\frac{2\sqrt{-C_1+u}}{\sqrt{1+4 C_1}}\right)=\ln\left(\frac{2\sqrt{-C_1+u}+\sqrt{4u+1}}{\sqrt{1+4C_1}}\right) \end{align*}
and \begin{align*} \arctan\left(\frac{\sqrt{-C_1+u}}{\sqrt{C_1+4C_1u}}\right)=\arcsin\left(\sqrt{\frac{-C_1+u}{u(1+4C_1)}}\right) \end{align*}
Therefore
\begin{align*} v=2\sqrt{C_1}\ln\left(\frac{2\sqrt{-C_1+u}+\sqrt{4u+1}}{\sqrt{1+4C_1}}\right)+\arcsin\left(\sqrt{\frac{-C_1+u}{u(1+4C_1)}}\right)+C_2 \end{align*}
Since $C_2$ is arbitrary and $\log a+\log b=\log(a\,b)$: \begin{align*} v&=2\sqrt{C_1}\ln\left(\frac{2\sqrt{-C_1+u}+\sqrt{4u+1}}{\sqrt{1+4C_1}}\right)+\arcsin\left(\sqrt{\frac{-C_1+u}{u(1+4C_1)}}\right)+\ln C_2 \\[12pt] v&=2\sqrt{C_1}\ln\left(C_2\,\frac{2\sqrt{-C_1+u}+\sqrt{4u+1}}{\sqrt{1+4C_1}}\right)+\arcsin\left(\sqrt{\frac{-C_1+u}{u(1+4C_1)}}\right) \end{align*}
\begin{align*} \arcsin\left(\sqrt{\frac{-C_1+u}{u(1+4C_1)}}\right)=v-2\sqrt{C_1}\ln\left(C_2\,\frac{2\sqrt{-C_1+u}+\sqrt{4u+1}}{\sqrt{1+4C_1}}\right) \end{align*}
\begin{align*} \frac{-C_1+u}{u(1+4C_1)}=\sin^2\left(v-2\sqrt{C_1}\ln\left(C_2\,\frac{2\sqrt{-C_1+u}+\sqrt{4u+1}}{\sqrt{1+4C_1}}\right)\right) \end{align*}
\begin{align*} u-C_1=u(1+4C_1)\sin^2\left(v-2\sqrt{C_1}\ln\left(C_2\,\frac{2\sqrt{u-C_1}+\sqrt{4u+1}}{\sqrt{1+4C_1}}\right)\right) \end{align*}
which is almost identical to the textbook solution:
\begin{equation} u-c^2=u(1+4c^2)\sin^2\left(v-2c\ln\left(k\left(2\sqrt{u-c^2}+\sqrt{4u+1}\right)\right)\right) \end{equation}
How can I get rid of the denominator $\sqrt{1+4C_1}$?
Too long for a comment. It's cumbersome. Two identities can help.
Using the one relating $\arctan$ and $\arcsin$:
$\arctan\left(\dfrac{\sqrt{-C_1+u}}{\sqrt{C_1+4 C_1\,u}}\right)=\arcsin\left(\dfrac{\dfrac{\sqrt{-C_1+u}}{\sqrt{C_1+4 C_1\,u}}}{\sqrt{1+\left(\dfrac{\sqrt{-C_1+u}}{\sqrt{C_1+4 C_1\,u}}\right)^2}}\right)=$
$=\arcsin\left(\dfrac{\dfrac{\sqrt{-C_1+u}}{\sqrt{C_1+4 C_1\,u}}}{\sqrt{\dfrac{u+4C_1u}{C_1+4 C_1\,u}}}\right)=\arcsin\left(\sqrt{\dfrac{-C_1+u}{u(1+4C_1)}}\right)$
Compare with the rearranged from MathWorld
$\begin{equation} \sqrt{\dfrac{u-c^2}{u(1+4c^2)}}=\sin\left(v-2c\ln\left(k\left(2\sqrt{u-c^2}+\sqrt{4u+1}\right)\right)\right) \end{equation}$
Further, from an identity with $\text{arcsinh}$
$\text{arcsinh}\left(\dfrac{2\sqrt{C_1}\sqrt{-C_1+u}}{\sqrt{C_1(1+4 C_1)}}\right)=\text{arcsinh}\left(\dfrac{2\sqrt{-C_1+u}}{\sqrt{1+4 C_1}}\right)=$
$=\ln\left(\dfrac{2\sqrt{-C_1+u}}{\sqrt{1+4 C_1}}+\sqrt{\left(\dfrac{2\sqrt{-C_1+u}}{\sqrt{1+4 C_1}}\right)^2+1}\right)=$
$=\ln\left(\dfrac{2\sqrt{-C_1+u}+\sqrt{4u+1}}{\sqrt{1+4 C_1}}\right)$
I think this can help.