I found this integral here: https://www.12000.org/my_notes/ten_hard_integrals/index.htm
Sites like https://www.integral-calculator.com/, which I often find very helpful, did not successfully calculate the integral.
I have found a potential answer, that seems to be correct.
Using the formula
$$\int f(x)\,dx=x \cdot f(x)-\left[{\int f^{-1}(y)\,dy}\right]_{y=f(x)},$$
I calculated :
$$\int \arcsin(\sqrt{x+1}-\sqrt{x})\,dx = x \cdot \arcsin(\sqrt{x+1}-\sqrt{x})+\frac{3}{8}\arcsin(\sqrt{x+1}-\sqrt{x})+\frac{1}{8}\sqrt{\sqrt{x(x+1)}-x}\cdot[(1+2\sqrt{2})\sqrt{x+1}-(1-2\sqrt{2})\sqrt{x}]$$
My question is: How else could I integrate this function? Is there an easier way? I tried to substitute $x=tan(u)^2$ but it seems that the resulting integral is an extremely long combination of sines and cosines.
Thank you for your help
Additional notes on my reasoning:
The inverse function of $\arcsin(\sqrt{x+1}-\sqrt{x})$ is $\left(\frac{\sin^2 x-1}{2\sin x}\right)^2$
Integrating this function can be done by expanding the square :
$$\int \left(\frac{\sin^2 y-1}{2\sin y}\right)^2 dy=\int \frac{1}{4}(\sin^2 y-2+\csc^2 y)\,dy$$
After having calculated that integral, we replace y by $\arcsin(\sqrt{x+1}-\sqrt{x})$. After some simplification work, I get the result mentioned above.
I just wanted to add on. I am still working on this problem. I was thinking that we could build of the identities:
$$\sin(a-b) = \sin(a)cos(b) - \cos(a)\sin(b)$$ $$a-b = \arcsin(\sin(a)\cos(b) - \cos(a)\sin(b))$$.
We could find two functions that we can integrate, $a$ and $b$, where $\sin(a)\cos(b) = \sqrt{x+1}$ and $\cos(a)\sin(b) = \sqrt{x}.$ I was thinking probably letting $a$ and $b$ being some inverse trig functions and proceeding from there.