# Thread: Partial Fractions (I think)

1. ## 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?

2. Just need to know a few things mate.

1) Is this the equation?

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

2) Are a and b both constants?

3. Yes and yes

4. In which case, I'd consider looking at substitution rather than partial fractions because of that annoying square root.

Try substituting $\displaystyle u = a^2 - b^2x^2$ into the equation and integrating that way.

5. 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

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

then

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

$\displaystyle du=-2b^2xdx$

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

The integral then is

$\displaystyle -\frac{1}{2b^2}\int{\frac{du}{\sqrt{u}}}$

6. Thanks guys. Easy once you select the correct method. Practice makes perfect.