The method of partial fractions splits rational functions into a sum of functions which
are like those which were just done successfully. In using this method it is essential that
in the rational function the degree of the numerator is smaller than the degree of the
denominator. The following simple but important Lemma shows that you can always
reduce to this case.
Lemma 7.7.4Let f
(x)
and g
(x)
≠0 be polynomials. Then there exists a polynomial,q
(x)
such that
f (x) = q (x)g (x )+ r(x)
where the degree of r
(x)
is less than the degree of g
(x)
or r
(x)
= 0.
Proof:Consider the polynomials of the form f
(x)
− g
(x )
l
(x)
and out of all
these polynomials, pick one which has the smallest degree. This can be done
because every nonempty set of nonnegative integers has a smallest member (Why?
Try to prove this by induction.). Let this take place when l
(x)
= q_{1}
(x)
and
let
r(x) = f (x)− g(x)q1(x).
It is required to show degree of r
(x)
< degree of g
(x )
or else r
(x)
= 0.
Suppose f
(x)
− g
(x)
l
(x)
is never equal to zero for any l
(x)
. Then r
(x)
≠0. It is
required to show the degree of r
(x)
is smaller than the degree of g
(x)
. If this doesn’t
happen, then the degree of r ≥ the degree of g. Let
m
r (x ) = bmx + ⋅⋅⋅+ b1x+ b0
g (x ) = anxn + ⋅⋅⋅+ a1x + a0
where m ≥ n and b_{m} and a_{n} are nonzero. Then let r_{1}
r(x)
◜-----◞◟------◝ xm −nbm
r1(x) = f (x)− g(x)q1(x)−---an--g (x)
( xm−nb )
= f (x)− g(x) q1 (x) + -----m- ,
an
and this is not zero by the assumption that f
(x)
− g
(x)
l
(x)
is never equal to zero for
any l
(x )
yet has smaller degree than r
(x)
which is a contradiction to the choice of r
(x )
.
■
Corollary 7.7.5Let f
(x)
and g
(x)
be polynomials. Then there exists a polynomial,r
(x)
such that the degree of r
(x)
< degree of g
(x)
and a polynomial, q
(x)
suchthat
f-(x)= q (x)+ r(x).
g(x) g(x)
Here is an example where the degree of the numerator exceeds the degree of the
denominator.
Example 7.7.6Find∫
3x52+7-
x −1
dx.
In this case the degree of the numerator is larger than the degree of the denominator
and so long division must first be used. Thus
5
3x-+-7-= 3x3 + 3x + 7+-3x-
x2 − 1 x2 − 1
Recall the process of long division from elementary school. The only difference is that you
use x raised to powers rather than 10 raised to powers. I am reviewing the algorithm in
what follows. The first term on the top is 3x^{3} because 3x^{3} times x^{2} gives 3x^{5} which will
cancel the first term of the 3x^{5} + 0x^{4} + 0x^{3} + 0x^{2} + 0x + 7. Then you multiply the
x^{2} + 0x− 1 by the 3x^{3} and subtract. Then you do the same process on 3x^{3} + 0x^{2} + 0x. A
more careful presentation of this algorithm is in my pre calculus book published by
worldwide center of math.
In this problem, first check to see if the degree of the numerator in the integrand is
less than the degree of the denominator. In this case, this is so. If it is not so, use
long division to write the integrand as the sum of a polynomial with a rational
function in which the degree of the numerator is less than the degree of the
denominator. See the preceding corollary which guarantees this can be done. Now look
for a partial fractions expansion for the integrand which is in the following
form.
--a---+ --b--+ -c2x-+d---
2x + 3 x + 5 x + x+ 1
and try to find constants, a,b,c, and d so that the above rational functions sum to the
integrand. The reason cx + d is used in the numerator of the last expression is that
x^{2} + x + 1 cannot be factored using real polynomials. Thus the problem involves finding
a,b,c,d, such that
Now these are two polynomials which are supposed to be equal. Therefore, they
have the same coefficients. Multiplying the right side out and collecting the
terms,
The solution is c = 1,a = 1,b = −2,d = 1. Therefore,
− x3 + 11x2 + 24x+ 14 1 2 1+ x
---------------2--------= ------− -----+ --2------.
(2x + 3)(x+ 5)(x + x +1) 2x + 3 x + 5 x + x+ 1
This may look like a fairly formidable problem. In reality it is not that bad.
First let x = −5 in 7.7 and obtain a simple equation for finding b. Next let
x = −3∕2 to get a simple equation for a. This reduces the above system to a more
manageable size. Anyway, it is now possible to find an antiderivative of the given
function.
∫ 1 ∫ 2 ∫ 1+ x
-----dx − -----dx+ -2------dx.
2x+ 3 x + 5 x + x+ 1
Each of these indefinite integrals can be found using the techniques given above. Thus the
antiderivative is
1
-ln|2x+ 3|− 2ln|x+ 5|+
2
√ - (√ - )
1ln(x2 + x + 1)+ 1 3arctan --3(2x + 1) + C.
2 3 3
What is done when the factors are repeated?
Example 7.7.8Find∫
(x+32x)+2(7x+3)
dx.
First observe that the degree of the numerator is less than the degree of the
denominator. In this case the correct form of the partial fraction expansion
is
--a--- ---b--- ---c--
(x+ 2) + (x + 2)2 + (x+ 3).
The reason there are two terms devoted to
(x + 2)
is that this is squared. Computing the
constants yields
----3x-+2-7----= --1--2-+ --2--− --2--
(x + 2) (x+ 3) (x + 2) x + 2 x + 3
and therefore,
∫
----3x+-7----dx = −--1--+ 2 ln (x + 2)− 2ln (x+ 3)+ C.
(x+ 2)2(x+ 3) x + 2
Example 7.7.9Find the proper form for the partial fractions expansion of
First check to see if the degree of the numerator is smaller than the degree of the
denominator. Since this is the case, look for a partial fractions decomposition in the
following form.
A B D gx+ h
------+ ------2 + ------+ -2---.
(x+ 2) (x + 2) (x+ 1) x + 1
These examples illustrate what to do when using the method of partial fractions. You
first check to be sure the degree of the numerator is less than the degree of the
denominator. If this is not so, do a long division. Then you factor the denominator into a
product of factors, some linear of the form ax + b and others quadratic, ax^{2} + bx + c
which cannot be factored further. Next follow the procedure illustrated in the above
examples and summarized below.
Warning: When you use partial fractions, be sure you look for somethingwhich is of the right form. Otherwise you may be looking for something which is not
there. The rules are summarized next.