ยง8.3. Part 2
Now that you have learned the basics of partial fractions, there are 3 more
tricks
that you need to know. (Just three, I promise.) They are:
• what happens if you have a repeated linear factor in the denominator
• what happens if you have a quadratic factor in the denominator that you
cannot factorize
• what happens if you have an improper fraction (namely that the degree
of the polynomial in the numerator is bigger than or equal to that of the
denominator)
Let's deal with them one by one .
(1)
Factorize the denominator : we get x2 +2x+1 = (x+1)2.
Therefore the
fraction that we are dealing with is
This is tricky because the factor x + 1 appears twice in
the denominator:
whenever we do partial fraction for these quotients, we need to (and this is
the only little extra bit that we need to know) decompose the fraction not
only in terms of 
, but also in terms of
. In other words, what we
now do is the following:
Let
Then proceed just as usual: the right hand side is
Hence we get
i.e. A = 1 and B = −1. Hence
Integrating, we get
Remember how to compute 
? (Hint: use substitution .)
The only trick here now is that you have to get the form of the partial
fraction correct. You couldn't solve the problem at all if you didn't know
the partial fraction of 
is going to look
like
The rule is the following : whenever (2x + 3)2 appears in
the denominator,
use both 
and
; in other words, your partial fraction will
involve
terms like
Whenever (2x + 3)3 appears in the denominator, you use all
of
and
in other words, your partial fraction will look like
Of course there is nothing special about 2 and 3 in 2x+3;
any other numbers
would work.
(2) Try now
(3)
The denominator can be factorized: it was already
partially done for
you. All that you need to do is to factorize it completely: so you want
to factorize x2 − 2x + 1 into (x − 1)2 and the whole denominator is just
(x + 1)(x − 1)2. Now do partial fraction: you get terms like
and
(Why?)
Now let
Then the right hand side is equal to
So we get
Hence from the second equation C = 2A, and from the third
we have
B = A + C = A + 2A = 3A. Plugging into the first equation, A + 3A = 4,
so A = 1, and then B = 3A = 3, C = 2A = 2. It fol lows that
Integrating, we get
(4) Try now
(5) Joke time!
Q: How do you make one burn?
A: Differentiate a log re!
(6) Next, if there is something that you cannot factorize in the denominator:
actually you can always factorize things (at least in principle ) until you get
quadratic factors, and whenever you cannot factorize a quadratic polyno-
mial that appears in the denominator, one of those little formula in Section
