Page 24 - DMTH202_BASIC_MATHEMATICS_II

Unit 2: Integration by Partial Fraction

then the rational function is known as proper, or else, it is known as improper. The improper Notes
rational functions can be abridged to the proper rational functions by long division procedure.

!

Caution The degree of the numerator ought to be less than the degree of the denominator.

2.1.1 Idea of Method of Partial Fractions

Here we show general rational function P(x)/Q(x) as a sum of integrable terms.
Assume that the degree of P(x) is p and that of Q(x) is q. If p is at least q, we can hide P by Q
(through synthetic, i.e., long division) to attain a polynomial, D(x), and a remainder, R(x)/Q(x),
along with the degree, r, of R less than q.
The fraction R(x)/Q(x) then moves toward 0 as YxY augments in every path in the multifaceted plane.

Each polynomial can be factored into linear factors (if complex numbers can occur in the factors).
As a result we get:
m
m
x
Q ( ) (x q 1 ) (x q 2 )  (x q k ) m 3
1




2
th
where m is the multiplicity of the k root of Q(x); the sum’s of the m’s being q.
k
!
Caution If the rational function is improper, make use of “long division” of polynomials
to carve it as the sum of a polynomial and a proper rational function “remainder.”
2.1.2 The Partial Fraction Theorem
You can consider R(x) / Q(x) as a sum over the roots q of the terms of the form
j
)
a  a (x q  a (x q ) m j  1


0 j 1 j j ( j m  1) j
j
(x q j ) m j

for suitable constants a .
jk
Notes Observe that this theorem permits us to integrate P(X) / Q(x); we are required only
to integrate the polynomial D(x) and the different inverse powers appearing in this sum,
(presuming we can calculate the a here.)
jk
Proof of the Partial Fraction Theorem

Let Q(a) = 0 and the root a of this equation contains multiplicity k.
Next, we have Q(x) = (x – a) Z(x) and Z(a) = c for non-zero c.
k
Let consider further that R(a) = d for non-zero d. (or else we could factor (x – a) out of both R and
Q and lessen their degrees.)

R ( ) d  k R ( ) d
x
x

Then  (x a ) acts at worst like (u x a )  (k 1) at x = a, since the rational function 
Q ( ) c Z ( ) c
x
x
disappears at x = a, and must consequently have a factor (x – a) in it.
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