On Wolfram Alpha's website, I find
$$\int\sqrt{x-\sqrt{x^2+1}}dx=\frac23 \sqrt{x - \sqrt{1 + x^2}}( {2 x + \sqrt{1 + x^2}})+C$$
I am trying to get this answer by substitution by $x=\tan(\theta), y=\sqrt{x-\sqrt{x^2+1}},$ or $y={x-\sqrt{x^2+1}}$. But the integrated converted to more difficult form .
Leonardo says this will not be easy to solve. With $$ t=\sqrt{x-\sqrt{x^2+1}} $$ I get $$ \int \sqrt{x-\sqrt{x^2+1}} \;dx = -2\int\sqrt{\frac{(t^4-1)^2}{4t^2}+1}\;dt =-\int \left(t^2+\frac{1}{t^2}\right)\;dt $$ and that integral is easy. $$\begin{align*} -\int \left(t^2+\frac{1}{t^2}\right)\;dt &= \frac{-t^3}{3}+\frac{1}{t}+C \\ &= \frac{-\left( x-\sqrt {{x}^{2}+1} \right) ^{3/2}}{3}+{\frac {1}{\sqrt {x- \sqrt {{x}^{2}+1}}}}+C \\ &=\frac23 \sqrt{x - \sqrt{1 + x^2}}\big( {2 x + \sqrt{1 + x^2}}\big)+C \end{align*}$$