How do I integrate inverse root functions in calc 2?

[silverdiesel]

I am taking calc 2 and I think we just finished up all the different ways of integrating, yet I can't figure this seemingly very simple one out. Any help is greatly appriciated.

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

 

[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? :)

 

[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.

 

[silverdiesel]

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

 

[d_leet]

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

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

 

[VietDao29]

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.

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...

 

[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.

 

[silverdiesel]

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

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

 

[d_leet]

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.

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

 

[d_leet]

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

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.

 

[VietDao29]

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

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? :)

 

[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 appreciate it!

 

[Tom McCurdy]

bit drunk

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

 

[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}}

 

[Mark44]

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.

 

[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.