# HINTS ON MATH 102 HOMEWORK 8

**§****.8.3.
Part 1**

We understand that you may not have done algebra for a while , so you may face

some real difficulty when you are integrating rational functions . Below are a
few

examples of how you could deal with such integrals; in particular, focus on why

and how the rational functions are decomposed into partial fractions. They are

almost the same as the assigned homework problems, except that some numbers

are different. Hopefully that will give you a hint on how to solve the homework

problems.

(1)

First one should see that such a fraction is a rational
function, meaning

that both the numerator (1 in this case) and the denominator (x^{2} + 3x

in this case) are polynomials. This suggests that techniques in handling

rational functions should be used. There are two main techniques: the

first one being an reduction (say by completing the square ) to one of those

formula in Section 8.1 (or Handout 3), which is applicable whenever the

denominator cannot be factorized, and the second one being decomposition

into partial fractions, which is applicable whenever the denominator can be

factorized. So let's do this integral.

First, the denominator of the integral is x^{2} + 3x. Can it be factorized?

In other words, can you write it as product of other ( lower -degree) polyno-

mials? (If you are unfamiliar with factorization, stop here and browse the

corresponding webpage that I have posted about factorization of polyno -

mials to remind yourself of how that works; we will just need some simple

factorizations for the moment.)

Yes, we have x^{2} + 3x = x(x + 3). So we would like to use the partial

fraction technique. In other words, we want to write
as combinations

of and .
So what do we do?

Let

where A and B are constants (meaning that they are numbers
that does

not depend on x). We want to figure out what A and B are. Let's simplify

the right hand side and compare with the left -hand side. The right hand

side is

So to make it match with the left hand side, we need

Solving, we get

and

so

(This can also be checked directly if you just compute
what the right hand

side is.) From here we can calculate the given integral, because the right

hand side is easy to integrate:

This completes the solution.

(2) Try now

(3)

Again, this is a rational function. Let's check whether
the denominator can

be factorized:

Yes! x^{2} − x = x(x − 1). So we want to do partial fraction:

Let

(Why?) Then the right hand side is

so comparing coefficients , we get

i.e. A = −4, B = 5. So

Integrating, we get

(4) Try now

(5) So the success of the whole method relies on two
things: first that you can

factorize the denominator, and second that you can decompose the given

rational function into (simpler) partial fractions that you can handle. The

first bit could be hard if you haven't done it for a while. Here is a brief

review of what you should know:

• Factorizing out any common factors in x (e.g. x^{2} + 3x, x^{3} + x)

• Factorizing difference of squares (e.g. x^{2} − 4, x^{2} − 1, 4x^{2} − 9)

• Factorizing a general quadratic polynomial (e.g. x^{2} − 7x + 12, 4x^{2} −

4x + 1)

If you have no trouble factorizing any of these you should
basically be fine.

If you have difficulty with any of them, read the webpage about factorization

before you proceed.

(6) Now a slightly more difficult one:

First, factorize the denominator: x^{2} + 7x + 12 = (x + 3)(x
+ 4).

Next, decompose the given fraction into partial fractions: Let

The right hand side is

Comparing coefficients, we get

It's not hard to solve this pair of equations : in case you
have forgotten,

here is how to do it. By the first equation, A = 3−B, so we can substitute

this into the second one, and get

4(3 − B) + 3B = 10

(we can do this substitution because both equations are simultaneously

true). In any event, we get 12−B = 10, i.e. B = 2, so A = 3−B = 3−2 = 1,

or in other words the solution is A = 1 and B = 2. Hence now

Integrating, we get

(7) Try now

(8) If you have taken the pain to go through all the above
and reached here,

you probably deserve a break. Here is a joke:

Q: Why do you rarely find mathematicians spending time at the beach?

A: Because they have sine and cosine to get a tan and don't need the

sun!

Read on...

(9) Sometimes you have to factorize a little bit more:

Let's factorize the denominator: x^{3}−x = x(x^{2}−1). But wait,
this is not

the end yet! Because x^{2}−1 can be further factorized: x^{2}−1 = (x−1)(x+1).

So altogether,

x^{3} − x = x(x − 1)(x + 1).

The game is that you will always have to factorize as
completely as possible

before you do partial fractions.

Now we do partial fractions: Let

Then the right hand side is equal to

So we get

Hence A = −1, and

Solving, we get B = 2 and C = −1. So

Integrating, we get

(10) Try now

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