# need help with this integral

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• Mar 24th 2011, 06:12 AM
gmarrero3
need help with this integral
I'm doing this long physics problem and I'm having problems with this integral, if someone could please help

int(dx/sqrt((1/a)-(1/x)))
• Mar 24th 2011, 06:27 AM
TheChaz
Are you allowed to use software? A certain popular online "calculator" gives a result that's too long for me to type or derive.
• Mar 24th 2011, 10:22 AM
SammyS
Re: need help with this integral
Quote:

Originally Posted by gmarrero3
I'm doing this long physics problem and I'm having problems with this integral, if someone could please help

$\displaystyle \int{{dx}\over{\sqrt{(1/a)-(1/x)}}}$

Added in Edit: IGNORE THIS POST !!
Multiply the integrand by:   $\displaystyle {{|ax|}\over{|ax|}}}={{|ax|}\over{\sqrt{a^2x^2}}}}$

This gives:  $\displaystyle \int{{|ax|}\over{\sqrt{x-a}}}\,dx\,.$

You'll need to decide how the absolute value affects your specific problem.

To evaluate the integral:   $\displaystyle a\int{{x}\over{\sqrt{x-a}}}\,dx\,,$ use integration by parts with:

$u = x$ and  $\displaystyle dv ={{dx}\over{\sqrt{x-a}}}$

Added in Edit: IGNORE THIS POST !!
• Mar 24th 2011, 10:35 AM
topsquark
Quote:

Originally Posted by gmarrero3
I'm doing this long physics problem and I'm having problems with this integral, if someone could please help

int(dx/sqrt((1/a)-(1/x)))

Where did you run into this integral? That might make the solution a bit easier to handle. It's doable without WolframAlpha (or some other program), but the solution is very picky and very tedious. On the other hand, it's not too hard to come up with the steps. I'll give you a quick run-down of what to do and give you the link to the details.

First get rid of the complex fractions in the integrand. That gives
$\displaystyle \int \frac{dx}{\sqrt{\frac{1}{a} - \frac{1}{x}}} = \int \sqrt{\frac{ax}{x - a}}~dx$

Now make the substitution x = ay. This gives (after some work)
$\displaystyle \int \sqrt{\frac{ax}{x - a}}~dx = a\sqrt{a} \int \sqrt{\frac{y}{y - 1}}~dy$

This integral looks not too bad, but as I said, it's a picky one. I'll leave the details to WolframAlpha, but it's really not that bad, just tedious.

The procedure is simple enough. First let $\displaystyle u = \frac{y}{y - 1}$. After you've simplified that, then let $s = \sqrt{u}$. That gives:
$\displaystyle \int \sqrt{\frac{y}{y - 1}}~dy = -2 \int \frac{s^2}{(s^2 - 1)^2}~ds$

which you can solve by partial fractions. Now backsubstitute to get
$\displaystyle \int \frac{dx}{\sqrt{\frac{1}{a} - \frac{1}{x}}} = \frac{a\sqrt{a}}{2} \left [ ln \left | \frac{\sqrt{x} + \sqrt{x - a}}{\sqrt{x} - \sqrt{x - a}} \right | + \left ( \frac{\sqrt{x - a}}{\sqrt{x} - \sqrt{x - a}} + \frac{\sqrt{x - a}}{\sqrt{x} + \sqrt{x - a}} \right ) \right ]$

As I said, without the physical problem I can't tell you what form of the solution might be the most useful.

-Dan
• Mar 24th 2011, 10:39 AM
topsquark
Quote:

Originally Posted by SammyS
Multiply the integrand by:   $\displaystyle {{|ax|}\over{|ax|}}}={{|ax|}\over{\sqrt{a^2 x^2}}}}$

This gives:  $\displaystyle \int{{|ax|}\over{\sqrt{x-a}}}\,dx\,.$

Watch that square root!

-Dan
• Mar 24th 2011, 06:13 PM
SammyS
Quote:

Originally Posted by topsquark
Watch that square root!

-Dan

Yup!

That sure was dumb of me !!!!
• Mar 24th 2011, 06:47 PM
topsquark
Quote:

Originally Posted by SammyS
Yup!

That sure was dumb of me !!!!

Mistakes happen. That does not make you dumb! Don't demean yourself unnecessarily.

-Dan