Answer
$$\sqrt{a(a+1)}+\ln|\sqrt{a}+\sqrt{a+1}|$$
Work Step by Step
Given
$$y= 2\sqrt{x} ,\ \ \ \ \ \ [0,a]$$
Since
\begin{align*}
y'&= \frac{1}{\sqrt{x}}\\
1+y'^2&= 1+ \frac{1}{x}
\end{align*}
Then the arc length is given by
\begin{aligned}
\text{Arc Length }&=\int_{a}^{b} \sqrt{1+\left(y^{\prime}\right)^{2}} d x \\
&=\int_{0}^{a} \sqrt{1+ \frac{1}{x}} d x \\
&= \int_{0}^{a} \sqrt{ \frac{x+1}{x}} d x \\
\end{aligned}
Use integration by parts
\begin{align*}
u&= \sqrt{ \frac{x+1}{x}} \ \ \ \ \ \ \ \ \ dv=dx\\
du &= -\frac{1}{2x^{\frac{3}{2}}\sqrt{x+1}}\ \ \ \ \ \ v= x
\end{align*}
Then
\begin{aligned}
\text{Arc Length } &= \int_{0}^{a} \sqrt{ \frac{x+1}{x}} d x \\
&= x \sqrt{ \frac{x+1}{x}} +\int \frac{dx}{2x^{\frac{1}{2}}\sqrt{x+1}}
\end{aligned}
Let $u=\sqrt{x},\ \ \ du=\frac{1}{2\sqrt{x}}$, then
\begin{align*}
\int \frac{dx}{2x^{\frac{1}{2}}\sqrt{x+1}}&= \frac{1}{2}\cdot \int \frac{2}{\sqrt{u^2+1}}du\\
&=\sinh^{-1}u\\
&=\ln \left|\sqrt{u^2+1}+u\right|\\
&=\ln \left|\sqrt{x+1}+\sqrt{x}\right|
\end{align*}
Hence \begin{aligned}
\text{Arc Length } &= \int_{0}^{a} \sqrt{ \frac{x+1}{x}} d x \\
&= \sqrt{ x(x+1)} + \ln \left|\sqrt{x+1}+\sqrt{x}\right|\bigg|_{0}^{a}\\
&=\sqrt{a(a+1)}+\ln|\sqrt{a}+\sqrt{a+1}|
\end{aligned}
