# $\int \frac{1}{1-\sqrt x} \mathrm dx$ 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?

-
+1 for showing work and initiative. – Arjang Mar 6 '13 at 21:43
Looks almost OK. Should be $\ln(|1-\sqrt{x}|)$. Importantly, you are missing the $+C$. Note that $2(1-\sqrt{x})$ can be replaced by $-2\sqrt{x}$. – André Nicolas Mar 6 '13 at 21:47
@milo In case it isn't clear from André's answer, you basically have the right answer. However, you can lump the $2$ from the $2(1-\sqrt{x})$ into the arbitrary constant. (and you should have abs signs in the log...) – apnorton Mar 6 '13 at 21:50
Would you provide us with results you're comparing with? – Kaster Mar 6 '13 at 22:03
I'm comparing it with $-2 (\sqrt[]x+ln|-1+\sqrt[]x|)$ which gives back the function if you take the derivative Considering what you guys told me about the constant I get $-2 (\sqrt[]x+\ln|1-\sqrt[]x|)$ – milo Mar 6 '13 at 22:27

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.

-
Old but great answer and general advice for working with sample solutions. – Nhat Aug 12 '14 at 22:35