Irrational integral $\int \frac{1}{\sqrt{x}} \sqrt{\frac{\sqrt{x}-2}{\sqrt{x}+2}}dx$ Would anyone be able to verify if this integral is calculated correctly?

$$\int \frac{1}{\sqrt{x}} \sqrt{\frac{\sqrt{x}-2}{\sqrt{x}+2}}dx$$

My attempt:
substitute:$\left(t = \sqrt{x}, t^2 = x, 2tdt=dx \right)$
$$
\begin{split}
\int \frac{1}{t} \sqrt{\frac{t-2}{t+2}}\,2t\,dt 
  &= 2\int\sqrt{\frac{t-2}{t+2}}dt
   = 2\int \frac{\sqrt{t-2}}{\sqrt{t+2}}dt \\
  &= 2\int \frac{\sqrt{(t-2)(t-2)}}{\sqrt{(t+2)(t-2)}}dt \\
  &= 2\int \frac{t-2}{\sqrt{t^2-4}}dt \\
  &= \int{\frac{2t-4}{\sqrt{t^2-4}}}dt \\
  &= \int \frac{2t}{\sqrt{t^2-4}}dt - 4\int\frac{dt}{\sqrt{t^2-4}} \\
  &= \sqrt{t^2-4} - 4\ln{|t+\sqrt{t^2-4}|} + C
\end{split}
$$
Substitute back $t = \sqrt{x}$:
result: $\Longrightarrow \sqrt{x-4} - 4\ln{|\sqrt{x} + \sqrt{x-4}|} + C$

$$...= \color{red}{2}\sqrt{t^2-4} - 4\ln{|t+\sqrt{t^2-4}|} + C$$
Anyway, you could take a derivate and check it you self.
