[m]\sqrt{x}=t[/m] ⇒
[m]x=t^2[/m]
[m]dx=d(t^2)=(t^2)`dt=2tdt[/m]
[m] ∫ \frac{dx}{\sqrt{x}+1}= ∫ \frac{2tdt}{t+1}= ∫ \frac{(2t+2)-2}{t+1}dt= ∫ (2-\frac{2}{t+1})dt=2t-2ln|t+1|+C=[/m]
[m]=2\sqrt{x}-2ln|\sqrt{x}+1|+C=2\sqrt{x}-2ln(\sqrt{x}+1)+C[/m]