Substitute [tex]y = \sqrt x[/tex], so that [tex]y^2 = x[/tex] and [tex]2y\,dy = dx[/tex]. Then the integral becomes
[tex]\displaystyle \int \frac{dx}{\sqrt{1 + \sqrt x}} = 2 \int \frac y{\sqrt{1+y}} \, dy[/tex]
Now substitute [tex]z=1+y[/tex], so [tex]dz=dy[/tex]. The integral transforms to
[tex]\displaystyle 2 \int \frac y{\sqrt{1+y}} \, dy = 2 \int \frac{z-1}{\sqrt z} \, dz = 2 \int \left(\sqrt z - \frac1{\sqrt z}\right) \, dz[/tex]
The rest is trivial. By the power rule,
[tex]\displaystyle \int \left(\sqrt z - \frac1{\sqrt z}\right) \, dz = \frac23 z^{3/2} - 2z^{1/2} + C = \frac23 \sqrt z (z - 3) + C[/tex]
Put everything back in terms of [tex]y[/tex], then [tex]x[/tex] :
[tex]\displaystyle 2 \int \frac y{\sqrt{1+y}} \, dy = \frac43 \sqrt{1+y} (y - 2) + C[/tex]
[tex]\displaystyle \int \frac{dx}{\sqrt{1+\sqrt x}} = \boxed{\frac43 \sqrt{1+\sqrt x} (\sqrt x - 2) + C}[/tex]
