How to integrate $\int\frac{\sqrt x}{\sqrt {1-x}}dx$? I have $$\int\dfrac{\sqrt x}{\sqrt {1-x}}dx$$ I place $\sqrt x=t$ so I have $$\int\dfrac{2t^2}{\sqrt{1-t^2}}dt$$ then by parts I have $$2t^2\arcsin{t}-4\int t\arcsin t dt$$ Now, how can I proceed? If I still use by parts, I go back. 
 A: Putting $x=\sin^2 t$, we get:
$$\int \frac{\sqrt x}{\sqrt {1-x}}dx$$
$$=\int \frac{2\sin ^2 t\cos t dt}{\cos t}$$
$$=2\int \sin^2t dt$$
$$=\int (1-\cos 2t )dt$$
$$=t-\frac{1}{2}\sin 2t +C$$
You can substitute $t$ back
A: Hint:
The standard substitution for integrals of functions of the form $ f\Bigl(x,\sqrt{\frac{ax+b}{cx+d}}\Bigr)$, where $f(x,y)$ is a rational function of two variables , is to use the substitution $t=\sqrt{\frac{ax+b}{cx+d}}$. So here, one sets
$$t=\sqrt{\frac{x}{1-x}}\iff t^2=\frac{x}{1-x},\enspace t\ge 0.$$
The substitution results in
$$\int\dfrac{\sqrt x}{\sqrt {1-x}}\,\mathrm d\mkern1mu x=\int \frac{2t^2}{(1+t^2)^2}\,\mathrm d\mkern1mu t=2\biggl(\int \frac{1}{ 1+t^2 }\,\mathrm d\mkern1mu t-\int \frac{1}{(1+t^2)^2}\,\mathrm d\mkern1mu t\biggr)$$
Now $\displaystyle I_1=\int \frac{1}{ 1+t^2 }\,\mathrm d\mkern1mu t=\arctan t\;$ and $\;\displaystyle I_2=\int \frac{1}{(1+t^2)^2}\,\mathrm d\mkern1mu t\;$ is classically obtained from $I_1$ by an integration by parts.
A: Hint: Write ${2t^2}$   as   ${2t^2}-2+2$ before the step where you apply by parts
