How to integrate $\frac{\sqrt{x}}{1-\sqrt{x}}$? How to integrate $\frac{\sqrt{x}}{1-\sqrt{x}}$?
I tried by using integration by parts, but always got sucked. Should be very easy...
 A: Let $u = 1 -\sqrt x$, so that $$\mathrm du = -\frac 1 {2 \sqrt x}\,\mathrm dx \implies \mathrm dx = -2 \sqrt x\,\mathrm du = 2(u - 1)\,\mathrm du.$$ Then the problem is reduced to evaluating
$$\int \frac{\sqrt x }{1 - \sqrt x}\,\mathrm dx = \int \frac{1 - u}{u} \cdot 2(u - 1)\,\mathrm du.$$
Can you finish it from here?
A: Hint: Either amplify by the conjugate, or let $t=\sqrt x$.
A: Let $u=\sqrt{x}$ and $x=u^2$, $\frac{\sqrt{x}}{1-\sqrt{x}}=-1+\frac{1}{1-\sqrt{x}}=-1+\frac{1}{1-u}$, then
\begin{align}
\int{\frac{\sqrt{x}}{1-\sqrt{x}}dx}&=\int{\left(-1+\frac{1}{1-u}\right)(2udu)}\\
&=-\int{2udu}+2\int{\frac{u}{1-u}du} \\
&=-u^2+2\int{\left(-1+\frac{1}{1-u}\right)du} \\
&=-u^2-2u+2\ln|1-u|+C \\
&=-x-2\sqrt{x}+2\ln|1-\sqrt{x}|+C
\end{align}
A: $$\int \frac{ \sqrt{x}}{1 - \sqrt{x}} = - \int \frac{\sqrt{x}}{\sqrt{x}-1} = - \int \frac{ \sqrt{x}-1+1}{\sqrt{x}-1}= - \int dx - \int \frac{dx}{\sqrt{x}-1}$$
But, $ d( \sqrt{x} - 1) = \frac{1}{2 \sqrt{x} } dx $. Hence
$$\int \frac{dx}{\sqrt{x}-1} = 2 \int \frac{\sqrt{x} d( \sqrt{x}-1)}{\sqrt{x}-1}=2 \int \frac{(\sqrt{x}-1+1) d( \sqrt{x}-1)}{\sqrt{x}-1} = 2 \int d( \sqrt{x}-1) + 2 \int \frac{ d(\sqrt{x}-1)}{\sqrt{x}-1} = 2 (\sqrt{x}-1) + 2 \ln(\sqrt{x}-1)$$
A: Take $x=z^2$. This works several times when there is a root under the integral.
$x=z^2 \Rightarrow dx=2zdz$
$\int \frac{\sqrt x}{1-\sqrt x}dx=2\int \frac{z^2}{1-z}dz$
Now write $z^2$ as $z^2-1+1=(z-1)(z+1)-1$
$=2\int \frac{(z-1)(z+1)-1}{1-z}dz=-2\int ((z+1)-\frac{1}{1-z})dz=-(z+1)^2-2ln(1-z)+c$
