# How do you use partial fraction decomposition to decompose the fraction to integrate 2/((x-5)(x-3))?

Jun 21, 2015

First, separate it into the format:

$\frac{A}{x - 5} + \frac{B}{x - 3}$

Now cross-multiply and simplify:

$\frac{A \left(x - 3\right) + B \left(x - 5\right)}{\left(x - 5\right) \left(x - 3\right)}$

$= \frac{A x - 3 A + B x - 5 B}{\left(x - 5\right) \left(x - 3\right)}$

$= \frac{\left(A + B\right) x - 3 A - 5 B}{\left(x - 5\right) \left(x - 3\right)}$

Notice there's an $x$ term and an ${x}^{0}$ term (constants). We can equate them back to what the numerator originally was. There was no $x$ term originally, so:

$A + B = 0$
$A = - B$

$- 3 A - 5 B = 2$
$\implies 3 B - 5 B = 2$
$- 2 B = 2$
$\implies B = - 1$
$\implies A = 1$

You're basically done. I was taught that $\int \frac{1}{x} \mathrm{dx} = \ln | x | + C$... so:

$= \int \frac{1}{x - 5} - \frac{1}{x - 3} \mathrm{dx}$

$= \ln | x - 5 | - \ln | x - 3 | + C$

Wolfram Alpha doesn't seem to agree with me about absolute values here compared to here, so let's just try differentiating this to see what we get.

$\left[\frac{\mathrm{df} \left(x\right)}{\mathrm{dx}}\right] \text{ at } x > 0 = \frac{1}{x - 5} - \frac{1}{x - 3}$

$= \frac{\left(x - 3\right) - \left(x - 5\right)}{\left(x - 5\right) \left(x - 3\right)}$

$= \frac{x - 3 - x + 5}{\left(x - 5\right) \left(x - 3\right)}$

$= \frac{2}{\left(x - 5\right) \left(x - 3\right)}$

...Yup, it works. $\text{*shrug*}$