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

Question

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: $\Longrightarrow \sqrt{x-4} - 4\ln{|\sqrt{x} + \sqrt{x-4}|} + C$

Answer 1

$$...= \color{red}{2}\sqrt{t^2-4} - 4\ln{|t+\sqrt{t^2-4}|} + C$$

Anyway, you could take a derivate and check it you self.

Answer 2

I got $2\sqrt{x-4}-8\operatorname{arcsinh}\left(\frac{\sqrt{\sqrt{x}-2}}{2}\right)+C$