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May 20th

May 20th

# HINTS ON MATH 102 HOMEWORK 8

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