# Integration of inverse root

1. Feb 11, 2006

### silverdiesel

I am taking calc 2 and I think we just finished up all the different ways of integrating, yet I cant figure this seemingly very simple one out. :grumpy: Any help is greatly appriciated.

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

2. Feb 11, 2006

### VietDao29

Hint: You know this
$$\int x ^ {\alpha} dx = \frac{x ^ {\alpha + 1}}{\alpha + 1} + C, \quad \alpha \neq -1$$, right?
$$\int \sqrt{\frac{1}{x}} dx = \int \frac{dx}{\sqrt{x}} = \int x ^ {-\frac{1}{2}} dx = ?$$
Can you go from here? :)

3. Feb 11, 2006

### silverdiesel

sure, but when I try that, the answer does not work. That yeilds:
$$2\sqrt{x}$$

I know the area under the curve already, and when I use $$2\sqrt{x}$$ in the definate integral, it does not give the correct area.

4. Feb 11, 2006

### silverdiesel

I think the answer should be:
$$2x\sqrt{1/x}$$
but I am not sure how to get there.

5. Feb 11, 2006

### d_leet

Ummm... but that is equivalent to $$2\sqrt{x}$$

6. Feb 11, 2006

### VietDao29

May you tell me the whole problem? To me, that's correct.
Just a small error there: You forgot to add the constant of integration (i.e, + C) into your result...

7. Feb 11, 2006

### silverdiesel

although it is really quite confusing becuase when I use my calculator to integrate $$\int\sqrt{\frac{1}{x}}$$, it gives the $$2x\sqrt{\frac{1}{x}}$$. Yet, when I use the calculator to differentiate $$2x\sqrt{\frac{1}{x}}$$, it does not give me $$\int\sqrt{\frac{1}{x}}$$. Unless, it gives it in a more complicated form that I have yet to work back to the original.

8. Feb 11, 2006

### silverdiesel

can you explain? Could be that my algebra is lacking.

9. Feb 11, 2006

### d_leet

What are you talking about? Could you restate this a bit more clearly.

10. Feb 11, 2006

### d_leet

the square root of 1/x is the square root of 1 divided by the square root of x, which is equal to 1 over the square root of x, and x divided by the square root of x is the square root of x, multiply that times 2 and you get 2 times the square root of x.

11. Feb 11, 2006

### VietDao29

Okay, one suggestion, though. Please don't rely heavily on calculators, and you can try do it by pen and paper, right?
$$2x \sqrt{\frac{1}{x}} = 2(\sqrt{x}) ^ 2 \frac{1}{\sqrt{x}} = 2 \sqrt{x}$$. Assuming that x >= 0. Can you get this? :)

12. Feb 11, 2006

### silverdiesel

okay, nevermind,

$$2\sqrt{x}$$

does work in the definate integral... and so it is the correct solution. Maybe it is time for me to go to bed.

Thanks for the help, I really appriciate it!

13. Feb 11, 2006

### Tom McCurdy

bit drunk

bit drunk but
$$\int\sqrt\frac{1}{x}dx$$
equals
$$2 \sqrt{\frac{1}{x}}x$$

14. Nov 15, 2009

### FortranMan

by the way, what's the proof for this inverse root? I swear I knew it at one time.

$$\int \frac{dx}{(x^2 + a^2)^{3/2}}=\frac{x}{a^2 \sqrt{x^2 + a^2}}$$

15. Nov 15, 2009

### Staff: Mentor

I can't believe this thread went on for as long as it did without the OP realizing that the integrand could be greatly simplified.
$$\sqrt{\frac{1}{x}}~=~\frac{\sqrt{1}}{\sqrt{x}}~=~\frac{1}{x^{1/2}}~=~x^{-1/2}$$

So the antiderivative requires nothing more than an application of the power rule.

16. Nov 15, 2009

### silverdiesel

wow.

3 years later, and yes, I think I've finally figured it out.

So, I am sitting at my desk trying to wrap my mind around the Kroneker Delta and the Levi-Civita Symbol. I do a google search for info, and a thread on my old friend, the Physics Forums pops up. I click the link, read it, (it did not help), I move on. Now, a few hours later I get an email about a response to a thread I posted three years ago.

I am guessing PF recognized me and made my account active again.

Anyhow... yeah, I got that one figured out. Funny how it seems so simple once you "speak" math. But when your just learning, the leap from x/root(x) to root(x) can be evasive.