# Need Help with Complicated Algebra Transformation in an Integral

• Sep 22nd 2009, 07:57 AM
PD1337
Need Help with Complicated Algebra Transformation in an Integral
Today in math class my teacher performed a complicated transformation in order to solve an integral. He changed:

1/[(x-2)(x+2)]

to (1/4)/(x-2) - (1/4)/(x+2)

This transformation was necessary to solve the problem. He did not tell me however how he did this and left it up to me to find. Does anyone know how this was done?
• Sep 22nd 2009, 08:03 AM
Prove It
Quote:

Originally Posted by PD1337
Today in math class my teacher performed a complicated transformation in order to solve an integral. He changed:

1/[(x-2)(x+2)]

to (1/4)/(x-2) - (1/4)/(x+2)

This transformation was necessary to solve the problem. He did not tell me however how he did this and left it up to me to find. Does anyone know how this was done?

He used a Partial Fractions decomposition.

Recall from when you were doing fractions that

$\displaystyle \frac{a}{b} + \frac{c}{d} = \frac{ad + bc}{bd}$.

He has had to go from the right hand side to the left hand side.

We write

$\displaystyle \frac{1}{(x -2)(x + 2)} = \frac{A}{x - 2} + \frac{B}{x + 2}$

$\displaystyle \frac{1}{(x - 2)(x + 2)} = \frac{A(x + 2) + B(x - 2)}{(x - 2)(x + 2)}$.

Clearly the numerators are equal...

$\displaystyle A(x + 2) + B(x - 2) = 1$

$\displaystyle Ax + 2A + Bx - 2B = 1$

$\displaystyle (A + B)x + 2A - 2B = 0x + 1$.

Equating like powers of x gives

$\displaystyle A + B = 0$ and $\displaystyle 2A - 2B = 1$

$\displaystyle A = -B$

$\displaystyle 2(-B) - 2B = 1$

$\displaystyle -4B = 1$

$\displaystyle B = -\frac{1}{4}$.

$\displaystyle A = -B$

$\displaystyle A = \frac{1}{4}$.

Therefore, finally

$\displaystyle \frac{1}{(x - 2)(x + 2)} = \frac{\frac{1}{4}}{x - 2} - \frac{\frac{1}{4}}{x + 2}$.

Now it is easy to integrate.
• Sep 22nd 2009, 08:11 AM
PD1337
That's great. Thanks so much for your help.