Key: remember the two basic
patterns
You should
realize very quickly that partial fraction decomposition has absolutely
nothing to do with Calculus. Partial fractions allow you to break
apart fractions that are nearly impossible to integrate into smaller
fractions that are simple to integrate. So I guess you could say
Calculus involves partial fractions, but partial fractions don't
involve Calculus. This focuses on partial fractions. Just stick
an integral symbol in front of each term to make it focus on Calculus.
Suppose you have a fraction 1 / 15. 1 / 15 is equal to 1 / (3
* 5) which is equal to 1 / (3 * (3 + 2)). Partial fractions allow
us to say that
Now if you multiply across
by 3 * (3 + 2) you'll get 1 = A * (3 + 2) + B * 3. Now you must
solve for the constants A and B. And that's it; the partial fraction
decomposition is over.
Granted, it does get slightly more complicated when x and y are
involved. But there are a few guidelines to follow that make it
very easy.
Since you end up with a cubed term after decomposing, this rule isn't much help by itself. It helps a great deal when it is mixed with the first rule, however.
It is very important to realize how the two rules could be mixed together. Suppose you have two unlike terms being multiplied together. And each unlike term is raised to a power (ie. like terms being multiplied together). For example:
To solve for the constants,
simply mutiply across by b(x). That will make ALL the fractions
disappear. If the fractions don't go away, you've done something
incorrectly. Supposing you are OK so far, now set x equal to values
that make some (or one) of the terms shrink to zero. Then you
should get a simple equation like 1 = A * 0 + B * 5 + C * 0. Then
you can assume from there on that B is equal to 1 / 5. Now solve
for a different constant.
Now that you've broken down the original fraction into smaller
fractions, and solved for all the resulting constants, put the
constants back into the original partial fractions. You should
now find that integrating will be more normal.
Probable Question: "My a(x) is not equal to 1."
That's fine. Sometimes you will get things like x - 1 = B * (x
- 1) + C * x. But look what transpires if you let x = 1. You get
0 = B * 0 + C * 1. Personally, I find it easier when b(x) is not
equal to a constant.
Tip: integration with partial fraction decomposition usually leads
to natural logs and arctans (Tan-1).