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Class Notes Day 26 on Partial Fractions

Class Notes Day 26 on Partial Fractions

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math 131, techniques of integrati<strong>on</strong> v partial fracti<strong>on</strong>s 2<br />

∫<br />

∫<br />

∫<br />

1<br />

1 + x 2 dx, x 3x 2 ∫<br />

1 + x 2 dx, − 1<br />

x 3 − x dx, and 1<br />

dx. But what about very similar<br />

looking functi<strong>on</strong>s such as<br />

∫ ∫ 1 + x<br />

2x + 1<br />

x 3 − x dx or 4<br />

dx? Why are these integrals<br />

4 − x2 not so easy to do? Over the next few secti<strong>on</strong>s, we examine a series of special cases<br />

of rati<strong>on</strong>al functi<strong>on</strong>s that we will see are relatively easy to integrate using a technique<br />

known as partial fracti<strong>on</strong>s.<br />

7.1 Special Cases: Linear Factors with Degree p(x) < q(x)<br />

There are several techniques that can be used integrate rati<strong>on</strong>al functi<strong>on</strong>s. We will<br />

c<strong>on</strong>centrate <strong>on</strong> a single technique and a couple of simple variati<strong>on</strong>s that work with<br />

rati<strong>on</strong>al functi<strong>on</strong>s of a particular type.<br />

Assume we have a rati<strong>on</strong>al functi<strong>on</strong> of the form<br />

r(x) = p(x)<br />

q(x) ,<br />

where degree of p(x) < degree of q(x) and q(x) can be factored into linear factors<br />

(factors of degree 1). Some examples include<br />

• 2x + 1<br />

x 3 − x = 2x + 1<br />

x(x 2 − 1) = 2x + 1<br />

where all the factors in the denominator<br />

x(x − 1)(x + 1)<br />

are linear and different and degree p(x) = 1 and degree q(x) = 3.<br />

4<br />

•<br />

4 − x 2 = 4<br />

; the two factors in the denominator are linear and<br />

(2 − x)(2 + x)<br />

different and degree of p(x) < degree of q(x).<br />

6x − 2<br />

•<br />

x 2 − 2x − 3 = 6x − 2<br />

where all the factors in the denominator are linear<br />

(x − 3)(x + 1)<br />

and different and degree of p(x) < degree of q(x).<br />

• x + 4<br />

x 3 + x 2 = x + 4<br />

; the three factors in the denominator are linear and<br />

x · x(x + 1)<br />

degree of p(x) < degree of q(x).<br />

On the other hand the rati<strong>on</strong>al functi<strong>on</strong><br />

2x + 1<br />

x 3 + x = 2x + 1<br />

x(x 2 + 1)<br />

does not satisfy our criteri<strong>on</strong> above because the denominator c<strong>on</strong>tains a factor of<br />

degree 2, in particular, x 2 + 1 cannot be written as a product of linear factors.<br />

We will use a technique called partial fracti<strong>on</strong>s to integrate the special rati<strong>on</strong>al<br />

functi<strong>on</strong>s above. 2<br />

The Key to <strong>Partial</strong> Fracti<strong>on</strong>s<br />

2 <strong>Partial</strong> fracti<strong>on</strong>s can be used to integrate<br />

other types of rati<strong>on</strong>al functi<strong>on</strong>s.<br />

Your text has further examples.<br />

Let’s look at a couple of ordinary fracti<strong>on</strong>s and how they can be rewritten in ‘simpler’<br />

terms. Notice<br />

1<br />

20 = 1 4 − 1 5<br />

or<br />

10<br />

21 = 1 3 + 1 7 .

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