# Basic integration - 1/sqrt(x)

1. Oct 27, 2012

### DrOnline

1. The problem statement, all variables and given/known data
$$\int \frac{1}{\sqrt{x}}$$

2. Relevant equations
$$\int x^n = \frac{1}{(n+1)} * X^{(n+1)} + c$$
$$\int \frac{1}{x} = ln|x| + c$$

3. The attempt at a solution

Well I figured since $$\int \frac {1}{x} = ln|x| + c, then \int \frac {1}{\sqrt{x}} = ln|{\sqrt{x}}| + c$$

But that's wrong. Apparently I have to use the first rule I stated, and i becomes $$2x^{(1/2)}$$

Which I can accept and do in the future, but I want to know why the second rule does not apply?

Last edited: Oct 27, 2012
2. Oct 27, 2012

### sardonic

Your second integral in part 3 is incorrect. You should normally be able to integrate a function, then derive it to get the same function back. What happens if you take the derivative of $ln(\sqrt{x})$? Don't forget the Chain Rule!

3. Oct 27, 2012

### DrOnline

Uhm.. I get:

$$\frac {d}{dx} ln( \sqrt{x}) = \frac {1}{\sqrt{x}} * \frac {1}{2*{\sqrt{x}}} = \frac {1}{2x}$$

Which is.. not the same as the original.

Alright. I see it cannot be like I wanted it to be, but I still find it a bit.. counter intuitive, if that makes sense, because:

If:

$$\frac {d}{dx} ln(x) = \frac {1}{x},$$ for any value of X, if can be anything a ton of functions all nested together,

then surely:

$$\int \frac {1}{x} = ln(x)$$ for any x.

Such as $$\sqrt{x}$$

Last edited: Oct 27, 2012
4. Oct 27, 2012

### BloodyFrozen

$$\frac{1}{\sqrt{x}} = x^{\frac{-1}{2}}$$

Last edited by a moderator: Oct 27, 2012
5. Oct 27, 2012

### DrOnline

Yes, I am aware of that, but if we call $$\sqrt{x} = u$$

Then $$\int \frac {1} {u} = ln(u), or: ln(\sqrt{x})$$

Bu that is false, which to me seems to violate the rule.

Edit: fixed some faulty reasoning. Now my question is valid.

Last edited by a moderator: Oct 27, 2012
6. Oct 27, 2012

### Ray Vickson

You are confused because your are forgetting that to get 'ln', the thing inside the "d" needs to match the denominator. That is, if you have
$$\int \frac{d(\text{thing 1})}{\text{thing 2}}$$ you need thing 1 = thing 2. When you have
$$\int \frac{dx}{\sqrt{x}},$$ you are not matching up the denominator and the thing under the "d". You would get ln(√x) if you had
$$\int \frac{d \sqrt{x}}{\sqrt{x}}.$$
Basically, if you change variables to y = √x for the denominator, you need to use dy, not dx, in the numerator.

RGV

7. Oct 27, 2012

### DrOnline

Wow, that devastated my mind. I'm gonna chew on that one for a bit. I think I understand what you mean.

Thanks everybody!

8. Oct 27, 2012

### SammyS

Staff Emeritus
You are doing the method of substitution incorrectly.

By your argument, you could just as well have let $\displaystyle u=\frac{1}{\sqrt{x}}$ and come up with $\displaystyle \int u\,, \$ which would give a very different, and also incorrect, answer.

You may notice that in you textbook, there is always a differential, dx in the case of your original integral, included with the integration symbol.

So you must do more than substituting $\displaystyle u\ \ \text{for}\ \ \frac{1}{\sqrt{x}} \ .$ You must also substitute for dx.

In the case of $\displaystyle u=\sqrt{x}\,,\ \ du=\frac{1}{2\sqrt{x}}dx\ .$

$\displaystyle \int \frac{1}{\sqrt{x}}\,dx=\int \frac{2}{2\sqrt{x}}dx=\int 2\,du\ .$

Added in Edit: I see RGV beat me to it. We both agree, but expressed ourselves enough differently so that I'll leave my post here rather than deleting it. (I'm pretty slow at typing this stuff up.)

9. Oct 27, 2012

### Staff: Mentor

In addition to other comments already made in this thread, you are omitting the differential in each of your integrals. While it might not seem important at this stage in your learning, omitting the differential will come back to bite you in the butt in the very near future, especially when you start to work with trig substitutions.

The corrected versions of these integrals are:
$$\int \frac{dx}{\sqrt{x}}$$
$$\int x^n dx= \frac{1}{(n+1)} * x^{(n+1)} + c$$
$$\int \frac{dx}{x} = ln|x| + c$$

10. Oct 27, 2012

### DrOnline

I get it now. I never really considered the impact of the dx.

I am actually familiar with the substitution method, I got to 2nd order differential equations in the past, but I am kind of re-learning this stuff these days.

Thanks a lot everybody. Love this forum.