What is the primitive function of $\arcsin{\sqrt{x}}$ Any clues?
$$\int{\arcsin{\sqrt{x}}} \mathrm dx$$
 A: a) Integrate by parts with $u=\arcsin\sqrt x,\;\mathrm dv=\mathrm dx$, obtaining
$$\int{\arcsin{\sqrt{x}}}\, \mathrm dx=\frac1{2\sqrt{x(1-x)}}-\frac12\int\sqrt{\frac x{1-x}}\,\mathrm dx.$$
b) Now for a square root of a homographic function, use substitution:
$$t=\sqrt{\frac x{1-x}}\iff t^2=\frac x{1-x} \iff x=\frac{t^2}{1+t^2},\enspace t\ge 0$$
This yields, if I'm not mistaken,
$$\int\sqrt{\frac x{1-x}}\,\mathrm dx=2\int\frac{t^3\,\mathrm dt}{(1+t^2)^2}$$
which you integrate by partial fractions decomposition.
A: hint
You can put $x=\sin^2(t)$
and then integrate by parts.
A: Let $I=\int \arcsin(\sqrt {x})\,dx$.  Next, enforce the substitution $x=y^2$ so that 
$$I=\int 2y\arcsin(y)\,dy \tag 1$$
Integrating by parts the integral on the right-hand side of $(1)$ with $u=\arcsin(y)$ and $v=y^2$ reveals
$$\begin{align}
I&=y^2\arcsin(y)-\int \frac{y^2}{\sqrt{1-y^2}}\,dy\\\\
&=y^2\arcsin(y)-\int\frac{y^2-1+1}{\sqrt{1-y^2}}\,dy\\\\
&=y^2\arcsin(y)+\int \sqrt{1-y^2}\,dy-\int \frac{1}{\sqrt{1-y^2}}\,dy\\\\
&=(y^2-1)\arcsin(y)+\frac12y\sqrt{1-y^2}+\frac12\arcsin(y)+C\\\\
&=(x-1/2)\arcsin(\sqrt x)+\frac12\sqrt{x(1-x)}+C
\end{align}$$
