$\int \frac{1}{1-\sqrt x} \mathrm dx $ Evaluating I'm having problems with this integral. My attempt:
$$\int \frac{1}{1-\sqrt[]x} dx $$
$$1-\sqrt[]x=u$$
$$du= -\frac{1}{2 \sqrt[]x}dx \rightarrow 2(u-1)du=dx $$
$$2 \int \frac{u-1}{u}du = 2\int du - 2 \int \frac{1}{u}du= 2u - 2 \ln(u) = 2(1-\sqrt[]x)-2(\ln(1-\sqrt[]x)) $$
But the result is wrong.
Where's the mistake?
 A: Consistent with the comments below your question, let's put it all together. The first two three lines show your result, save for the missing arbitrary constant which you need to include:
$$2 \int \frac{u-1}{u}du \quad = \quad 2\int du - 2 \int \frac{1}{u}du $$ 
$$ = \quad 2u - 2 \ln(|u|) + c $$
$$ = 2(1-\sqrt[]x)-2(\ln(1-\sqrt[]x)) + c$$
$$= 2 - 2\sqrt x - 2\ln|1 - \sqrt x| + c $$
$$ = -2\sqrt x - 2 \ln|1 - \sqrt x| + C$$
Where we replace $c$ with $2 + c = C$, both $c$ and $C$ arbitrary constants.
Does this look closer to the (solution) result you are referring to? If so, you can see the expressions are equivalent solutions to the integral; and importantly, we have added "absolute value" signs surrounding the arguments of $\ln$. Recall that $$\int \frac 1u du = \ln|u| + C$$
One last observation: since we are working with $\ln|1 - \sqrt x|$ as a term in the solution, one can equivalently, express this as $\ln|\sqrt x - 1|$. That is, $$|1 - \sqrt x| = |\sqrt x - 1|,\quad \text{just as}\quad |1 - 3| = |3 - 1| = 2$$

This is a good example to keep in mind in the future:  although two solutions to an indefinite integration problem may appear different, depending on the substitution made, functions like the $\ln |u|$, or alternate trig substitutions may yield solutions that seem to differ, so you can have two, or even more, correct solutions, so long as they differ only in terms of a constant. You can always test a solution, if you're in doubt or if it doesn't match the answer key, by differentiating your solution. If that results in the original integrand, you've integrated successfully.
