1
Calculus 101 Section 205
Name/ID:
Partial fractions long example
February 2011
Guillermo Mantilla
The purpose of this note is to exhibit an example that shows how to find the antiderivative of a rational fucntion
by the method of partial fractions. The example I’ve chosen is very long and tedious but if you get through it
you’ll be in very good shape on this topic.
Before doing the example I’ll recall the necessary steps to integrate a rational function. Let Q(x), P (x) polynomials with real coefficients, and assume P (x) is non constant1
Problem : Find
they are.
R
Q(x)
P (x) dx.
There are four steps you need to follow to solve this problem, here I recall what
Setp 1: If degree(P (x)) ≤ degree(Q(x)) use long division to find polynomials Q1 (x) and R(x) with degree(P (x)) >
R Q(x)
R
R(x)
degree(R(x)), and such that Q(x)
Q1 (x)dx +
P (x) = Q1 (x) + P (x) . Once you have done this
P (x) dx =
R R(x)
R R(x)
P (x) dx, so the difficulty remains in finding
P (x) dx since Q1 (x) is a polynomial.
Setp 2: Factorize P (x) as a product of linear polynomials, meaning polynomials of degree 1, and irreducible
polynomials of degree 2. Recall that a polynomial ax2 + bx + c is irreducible if and only if b2 < 4ac. One
way to find linear factors of P (x) is a follows. Try to find values a such that P (a) = 0 (a is a root of
P (x)). If a is a root of P (x), then P (x) is divisible by x − a so using long division you can start factorizing
P (x).
R(x)
Setp 3: After you have factorized P (x) you write an equation of the form P
(x) = certainf ractions where in
“certain fractions” you put the following: If (x − a) is a linear factor of P (x) that appears repeated
A1
A2
An
exactly n-times in its factorization, then you add the term x−a
+ (x−a)
2 + ... + (x−a)n for some variables
A1 , ..., An . If ax2 + bx + c is an irreducible quadratic factor of P (x) that appears repeated exactly m-times
B1 x+C1
B2 x+C2
Bm x+Cm
in its factorization, then you add the term ax
2 +bx+c + (ax2 +bx+c)2 + ... + (ax2 +bx+c)m for some variables
B1 , ..., Bm , C1 , ..., Cm . After you add all the terms in “certain fractions” you’ll obtain a fraction which
numerator is a polynomial T (x) such that its coefficients depend linearly on the variables A0 s, B 0 s and
C 0 s... By comparing coefficients of T (x) and R(x) you obtain a k by k, k is degree(P (x)), system of linear
equations on the variables A0 s, B 0 s and C 0 s... that you must solve.
R R(x)
Setp 4: From the previous step you’ve reduced the problem of finding P
(x) dx to a sum of integrals of the form
R A
R
Bx+C
(x−a)i dx or
(ax2 +bx+c)j dx and that type of antiderivatives can be found fairly easy with methods
we’ve covered.
Remark: Even though Step 3 could seem long and tedious there is nothing complicated about it. You are just
adding, multiplying and comparing coefficients of polynomials. So as long P (x) has small degree that step is
actually very simple.(Normally you’ll have to deal only with P of degree at most 3.)
1
The long example
We want to find
Z x9 − x8 + 5x7 − 3x6 + 7x5 − 5x4 − x + 1
x8 − x7 + 2x6 − 2x5 + x4 − x3
dx
In the notation above we have that P (x) = x8 − x7 + 2x6 − 2x5 + x4 − x3 , and that Q(x) = x9 − x8 + 5x7 −
3x6 + 7x5 − 5x4 − x + 1.
Step 1 Since degree(P (x)) = 8 ≤ 9 =degree(Q(x)) we need to use long division, i.e. divide Q(x) by P (x). By
doing so
x9 − x8 + 5x7 − 3x6 + 7x5 − 5x4 − x + 1
3x7 − x6 + 6x5 − 4x4 − x + 1
=x+ 8
8
7
6
5
4
3
x − x + 2x − 2x + x − x
x − x7 + 2x6 − 2x5 + x4 − x3
i.e., Q1 (x) = x and R(x) = 3x7 − x6 + 6x5 − 4x4 − x + 1.
1 If
P (x) is constant the problem is trivial.
2
Step 2 We need to factorize x8 − x7 + 2x6 − 2x5 + x4 − x3 . Clearly x3 is a factor, so we write x8 − x7 + 2x6 − 2x5 +
x4 − x3 = x3 (x5 − x4 + 2x3 − 2x2 + x − 1). Since the remaining factor h(x) = x5 − x4 + 2x3 − 2x2 + x − 1
has degree 5 we have to keep factorizing. Since h(1) = 0 we know that x − 1 divides h(x). By doing long
division we get that h(x) = (x − 1)(x4 + 2x2 + 1). We are left with x4 + 2x2 + 1, and since this polynomial
has degree 4 we can factorize it more. In fact it is simple to see that x4 + 2x2 + 1 = (x2 + 1)2 . Since x2 + 1
is irreducible (0 < 4) we have completed the factorization of P (x).
Step 3 From Step 2 we conclude that x8 − x7 + 2x6 − 2x5 + x4 − x3 = x3 (x − 1)(x2 + 1)2 , in other words the
irreducible factors of P (x) are x repeated three times, (x − 1) repeated one time and x2 + 1 repeated twice.
Therefore we take unknowns A, B, C, D, E, F, G, H and set the equation
A
D
Ex + F
B
C
Gx + H
3x7 − x6 + 6x5 − 4x4 − x + 1
=
.
+ + 2+ 3+ 2
+ 2
x8 − x7 + 2x6 − 2x5 + x4 − x3
x−1
x
x
x
x +1
(x + 1)2
The least common multiple of x − 1, x, x2 , x3 , x2 + 1 and (x2 + 1)2 is (x − 1)(x3 )(x2 + 1)2 so the term in
the right hand side is a fraction with denominator (x − 1)(x3 )(x2 + 1)2 . Adding all the fractions on the
right hand side of the above equality one gets:
Ax3 (x2 +1)2 +B(x−1)x2 (x2 +1)2 +C(x−1)x(x2 +1)2 +D(x−1)(x2 +1)2 +(Ex+F )x3 (x−1)(x2 +1)+(Gx+H)x3 (x−1)
(x−1)x3 (x2 +1)2
=
A(x7 +2x5 +x3 )+B(x7 −x6 +2x5 −2x4 +x3 −x2 )+C(x6 −x5 +2x4 −2x3 +x2 −x)+D(x5 −x4 +2x3 −2x2 +x−1)+E(x7 −x6 +x5 −x4 )+F (x6 −x5 +x4 −x3 )+G(x5 −x4 )+H(x4 −x3 )
=
x3 (x−1)(x2 +1)2
=
(A+B+E)x7 +(−B+C−E+F )x6 +(2A+2B−C+D+E−F +G)x5 +(−2B+2C−D−E+F −G+H)x4 +(A+B−2C+2D−F −H)x3 +(−B+C−2D)x2 +(−C+D)x−D
.
x8 −x7 +2x6 −2x5 +x4 −x3
In the notation of the beginning
7
6
5
4
3
2
T (x) = (A+B+E)x +(−B+C−E+F )x +(2A+2B−C+D+E−F +G)x +(−2B+2C−D−E+F −G+H)x +(A+B−2C+2D−F −H)x +(−B+C−2D)x +(−C+D)x−D
Since T (x) must be equal to 3x7 − x6 + 6x5 − 4x4 − x + 1 the coefficient of each degree must be the same,
in other words
3=A+B+E
−1 = −B + C − E + F
(1)
(2)
6 = 2A + 2B − C + D + E − F + G
(3)
−4 = −2B + 2C − D − E + F − G + H
(4)
0 = A + B − 2C + 2D − F − H
(5)
0 = −B + C − 2D
(6)
−1 = −C + D
1 = −D.
(7)
(8)
From the equation (8) D = −1, so the equation (7) gives that C = 0. Putting that in equation (6) we see
that B = 2. By replacing the values we just found in the remaining equations we obtain:
1=A+E
(9)
1 = −E + F
(10)
3 = 2A + E − F + G
(11)
−1 = −E + F − G + H
(12)
0=A−F −H
(13)
From equations (12) and (13) we see that A = F + H and that E = 1 + F − G + H. By replacing these
in equations (9), (10), (11)
we obtain:
0 = 2F − G + 2H
(14)
2=G−H
(15)
2 = 2F + 3H
(16)
3
By adding the first two equations of the above system we obtain
2 = 2F + H
(17)
2 = 2F + 3H
(18)
from which immediately follows that H = 0, and consequently F = 1. Also, G = 2 by equation (15). It
follows from equations (12), (13) that A = 1 and E = 0.
If you know a bit of linear algebra then a faster and more systematic way to solve the system is using row
reduction. You can write the original system as


1 1
0
0
1
0
0
0 | 3
0 −1 1
0 −1 1
0
0 | −1


2 2 −1 1
1
−1
1
0 | 6


0 −2 2 −1 −1 1 −1 1 | −4


1 1 −2 2
0 −1 0 −1 | 0 


0 −1 1 −2 0
0
0
0 | 0


0 0 −1 1
0
0
0
0 | −1
0 0
0 −1 0
0
0
0 | 1
and then use Gauss-Jordan reduction

1
0

0

0

0

0

0
0
to get that the system is row equivalent to

0 0 0 0 0 0 0 | 1
1 0 0 0 0 0 0 | −2

0 1 0 0 0 0 0 | 0

0 0 1 0 0 0 0 | −1

0 0 0 1 0 0 0 | 0

0 0 0 0 1 0 0 | 1

0 0 0 0 0 1 0 | −2
0 0 0 0 0 0 1 | 0
which also leads us to A = 1, B = 2, C = 0, D = −1, E = 0, F = 1, G = 2, H = 0.
Step 4 Recall from Step 1 that
Z
x9 − x8 + 5x7 − 3x6 + 7x5 − 5x4 − x + 1
dx =
x8 − x7 + 2x6 − 2x5 + x4 − x3
Z
Z
xdx +
3x7 − x6 + 6x5 − 4x4 − x + 1
dx.
x8 − x7 + 2x6 − 2x5 + x4 − x3
From Steps 2 and 3
3x7 − x6 + 6x5 − 4x4 − x + 1
1
2
1
1
2x
=
+ −
+ 2
+
.
x8 − x7 + 2x6 − 2x5 + x4 − x3
x − 1 x x3
x + 1 (x2 + 1)2
Therefore
Z
x9 − x8 + 5x7 − 3x6 + 7x5 − 5x4 − x + 1
dx =
x8 − x7 + 2x6 − 2x5 + x4 − x3
Z
Z
xdx+
Z
1
dx+
x−1
2
dx−
x
Z
1
dx+
x3
Z
1
dx+
2
x +1
Z
Since we know how to find each one of the above antiderivatives we finally get:
Z
x9 − x8 + 5x7 − 3x6 + 7x5 − 5x4 − x + 1
x2
1
1
dx
=
+ ln |x − 1| + 2 ln |x| + 2 + tan−1 (x) − 2
+c
8
7
6
5
4
3
x − x + 2x − 2x + x − x
2
2x
x +1
Note: It took SAGE (http://www.sagemath.org/) 0.04 seconds on a standard laptop to do the above antiderivative.
(x2
2x
dx.
+ 1)2