How to integrate $\int\frac{\sqrt x}{\sqrt {1-x}}dx$?

Question

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.

Answer 1

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

Answer 2

Hint: Write ${2t^2}$ as ${2t^2}-2+2$ before the step where you apply by parts

Answer 3

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.