# Partial Fractions (I think)

• Mar 4th 2010, 02:55 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?
• Mar 4th 2010, 03:04 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?
• Mar 4th 2010, 03:08 AM
p75213
Yes and yes
• Mar 4th 2010, 03:16 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.
• Mar 4th 2010, 03:16 AM
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

$-\frac{1}{2b^2}\int{\frac{du}{\sqrt{u}}}$
• Mar 4th 2010, 03:34 AM
p75213
Thanks guys. Easy once you select the correct method. Practice makes perfect.