# Partial Fractions (I think)

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• March 4th 2010, 01:55 AM
p75213
Partial Fractions (I think)
! = integrate
I am trying to integrate the following:
! x/sqr(a^2-b^2*x^2) dx
Using partial fractions I get the following:
1/sqr(2b) ! sqr(x)/sqr(a+bx) dx - 1/sqr(2b) ! sqr(x)/sqr(a-bx) dx.
This doesn't seem to help at all so it has obviously come off the rails. Could someone be so kind as to show the correct workings?
• March 4th 2010, 02:04 AM
Gojinn
Just need to know a few things mate.

1) Is this the equation?

$\dfrac{x}{\sqrt{a^2-b^2x^2}}$

2) Are a and b both constants?
• March 4th 2010, 02:08 AM
p75213
Yes and yes
• March 4th 2010, 02:16 AM
Gojinn
In which case, I'd consider looking at substitution rather than partial fractions because of that annoying square root.

Try substituting $u = a^2 - b^2x^2$ into the equation and integrating that way.
• March 4th 2010, 02:16 AM
Archie Meade
Quote:

Originally Posted by p75213
! = integrate
I am trying to integrate the following:
! x/sqr(a^2-b^2*x^2) dx
Using partial fractions I get the following:
1/sqr(2b) ! sqr(x)/sqr(a+bx) dx - 1/sqr(2b) ! sqr(x)/sqr(a-bx) dx.
This doesn't seem to help at all so it has obviously come off the rails. Could someone be so kind as to show the correct workings?

If you use the substitution

$u=a^2-b^2x^2$

then

$\frac{du}{dx}=-2b^2x$

$du=-2b^2xdx$

$xdx=-\frac{du}{2b^2}$

The integral then is

$-\frac{1}{2b^2}\int{\frac{du}{\sqrt{u}}}$
• March 4th 2010, 02:34 AM
p75213
Thanks guys. Easy once you select the correct method. Practice makes perfect.

Answer is -sqr(a^2-b^2*x^2)/b^2.

Gojinn: Your right about the ANNOYING square root.
• March 4th 2010, 03:02 AM
Gojinn
Yeah, those buggers can really make life more tedious. I mean, at first it's kinda handy because if, in my experience, if you see a big square root in the denominator then you'll probably have to use substitution... but once you're over the joy of knowing which technique to use, you realise the extra work you gotta do to find the answer. :P

Good luck with you other integrals. ^_^