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$$\int \frac{dx}{x\sqrt{1-x}}$$

$$\int \frac{dx}{x\sqrt{1-x}}$$

$u=1-x$

$du=-dx$

$$-\int \frac{du}{(1-u)\sqrt{u}}$$

$a(1-u)+b\sqrt{u}=1\Rightarrow a-au+b\sqrt{u}=1$

$a=1\Rightarrow b\sqrt{u}-u=0\Rightarrow b=\sqrt{u}$

$$\int \frac{\sqrt{u}}{1-u}du-\int \frac{du}{\sqrt{u}}=\int \frac{\sqrt{u}}{1-u}du-2{\sqrt{u}}$$

$$\int \frac{\sqrt{u}}{1-u}du=\int \frac{1+\sqrt{u}-1}{(1+\sqrt{u})(1-\sqrt{u})}du=\int \frac{1+\sqrt{u}}{(1+\sqrt{u})(1-\sqrt{u})}du-\int \frac{du}{(1+\sqrt{u})(1-\sqrt{u})}du=\int \frac{1}{(1-\sqrt{u})}du-\int \frac{du}{(1+\sqrt{u})(1-\sqrt{u})}du$$

$$\int \frac{1}{(1-\sqrt{u})}du$$

How do I continue from here? it seems that I have made it harder

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    $\begingroup$ At $- \int \frac{du}{(1-u)\sqrt{u}}$, it may help to substitute $t = \sqrt{u}$. $\endgroup$ – Daniel Fischer Feb 12 '16 at 13:21
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    $\begingroup$ The substitution you made seems wrong. $$\frac{1}{x\sqrt{1-x}}=\frac{1}{\sqrt u-u\sqrt u}$$ $\endgroup$ – user228113 Feb 12 '16 at 13:22
  • $\begingroup$ the denuminator of your third integral should read $\sqrt u - u \sqrt u$. $\endgroup$ – Math-fun Feb 12 '16 at 13:29
  • $\begingroup$ fixed the denuminator $\endgroup$ – gbox Feb 12 '16 at 13:35
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Set $t=\sqrt{1-x}$, so $x=1-t^2$ and $dx=-2t\,dt$, so the integral becomes $$ \int\frac{-2t}{(1-t^2)t}\,dt=\int\frac{2}{t^2-1}\,dt= \int\left(\frac{1}{t-1}-\frac{1}{t+1}\right)\,dt $$

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Hints:

1. Substitute, as you did, $u = 1-x$, $\text{d}u = -\text{d}x$

2. you get $\int\frac{1}{(u-1)\sqrt{u}}\ \text{d}u$

3. Substitute now $y = \sqrt{u}$, $\text{d}y = \frac{1}{2\sqrt{u}}\ \text{d}u$

4. Integral is now $-2\int\frac{1}{1 - y^2}\ \text{d}y$

5. Solution of that integration is $-2\ \text{arctanh}(y) + C$

Now you can substitute back to $x$ and get your answer

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  • $\begingroup$ yes it is $arctanh$, but I prefer $\frac{1}{2a}[ln(x-a)-ln(x+a)]+C$ $\endgroup$ – gbox Feb 12 '16 at 13:34
  • $\begingroup$ It was a typo! Sorry. $\endgroup$ – Von Neumann Feb 12 '16 at 14:17
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A yet different approach: \begin{align} I&=\int \frac{dx}{x\sqrt{1-x}}\\ &=\int \frac{-2t^{-2}+2t^{-3}}{(2t^{-1}-t^{-2})\sqrt{1-2t^{-1}+t^{-2}}}dt \mbox{, where $x=2t^{-1}-t^{-2}$}\\ &=\int \frac{-2t^{-2}(1-t^{-1})}{(2t^{-1}-t^{-2})(1-t^{-1})}dt \\ &=-2\int \frac{1}{2t-1}dt \\ &=-\ln(2t-1)\\ &=\ln\frac{x}{2+2\sqrt{1-x}-x} \mbox{, where $t=\frac{1+\sqrt{1-x}}{x}$}\\ &=\ln\frac{(1-\sqrt{1-x})(1+\sqrt{1-x})}{(1+\sqrt{1-x})^2}\\ &=\ln\frac{1-\sqrt{1-x}}{1+\sqrt{1-x}}\\ \end{align}

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HINT:

Let $ x = u^2, \, dx= 2 u du ; \, \int \dfrac{2 du }{u\sqrt{1-u^2} }$

again let $$ u = \sin(t) ; \ du = \cos (t) dt ...$$

leading to

$$ =2 log \dfrac{u}{1+\sqrt{1-u^2}} = 2 log \dfrac{\sqrt x}{1+\sqrt{1-x} }$$

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Hint 1:- Let, $\sqrt{u}=t\implies\dfrac{du}{\sqrt{u}}=2dt$. So, $$\int\dfrac{du}{(1-u)\sqrt{u}}=\int\dfrac{du}{(1-\sqrt{u})(1+\sqrt{u})\sqrt{u}}=2\int\dfrac{dt}{(1-t)(1+t)}$$

Hint 2:- $$I=\int\dfrac{du}{(1-u)\sqrt{u}}=\int\dfrac{du}{(1-\sqrt{u})(1+\sqrt{u})\sqrt{u}}=\int\dfrac{2(1-\sqrt{u})+(1+\sqrt{u})+\sqrt{u}}{(1-\sqrt{u})(1+\sqrt{u})\sqrt{u}}$$Hence we get, $$I=2\color{red}{\int\dfrac{du}{(1+\sqrt{u})\sqrt{u}}}+\color{blue}{\int\dfrac{du}{(1-\sqrt{u})\sqrt{u}}}+\color{green}{\int\dfrac{du}{(1-\sqrt{u})(1+\sqrt{u})}}=2\color{red}{I_1}+\color{blue}{I_2}+\color{green}{I_3}$$Further, $$\color{red}{I_1=\int\dfrac{du}{(1+\sqrt{u})\sqrt{u}}=\int\dfrac{(1+\sqrt{u})-\sqrt{u}}{(1+\sqrt{u})\sqrt{u}}du=\int\dfrac{du}{\sqrt{u}}-\int\dfrac{du}{1+\sqrt{u}}}$$$$\color{blue}{I_2=\int\dfrac{du}{(1-\sqrt{u})\sqrt{u}}=\int\dfrac{(1-\sqrt{u})+\sqrt{u}}{(1-\sqrt{u})\sqrt{u}}du=\int\dfrac{du}{\sqrt{u}}+\int\dfrac{du}{1-\sqrt{u}}}$$$$\color{green}{I_3=\int\dfrac{du}{(1+\sqrt{u})(1-\sqrt{u})}=\dfrac{1}{2}\int\dfrac{(1+\sqrt{u})+(1-\sqrt{u})}{(1+\sqrt{u})(1-\sqrt{u})}du=\int\dfrac{du}{1+\sqrt{u}}+\int\dfrac{du}{1-\sqrt{u}}}$$

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$x=\sin^2(\theta)$ $$ \begin{align} \int\frac{\mathrm{d}x}{x\sqrt{1-x}} &=\int\frac{2\,\mathrm{d}\theta}{\sin(\theta)}\\ &=-2\int\frac{\mathrm{d}\cos(\theta)}{1-\cos^2(\theta)}\\ &=-\int\left(\frac1{1-\cos(\theta)}+\frac1{1+\cos(\theta)}\right)\mathrm{d}\cos(\theta)\\ &=\log\left(\frac{1-\cos(\theta)}{1+\cos(\theta)}\right)+C\\ &=\log\left(\frac{1-\sqrt{1-x}}{1+\sqrt{1-x}}\right)+C\\ \end{align} $$

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