Integration of Polynomial Rational Expressions
Partial Fractions
© 2010 Scott Surgent
A polynomial rational expression is of the form
, where both and are polynomials. It is
considered “proper” if the degree of the top is less than the degree of the bottom. If the degree of the top is
equal to or greater than that of the bottom, the expression is considered “improper”.
The goal is to decompose the expression
into the sum of a series of smaller rational expressions, whose
denominators are either linear or irreducible quadratics. A corollary of the Fundamental Theorem of Algebra
states that any polynomial with Real coefficients can be factored into a product of linear and/or irreducible
quadratic factors. A linear factor is of the form + , and an irreducible quadratic is of the form
+ + such that it has complex roots.
There are many cases to consider:
Case I: The polynomial factors into linear factors, none of which appears more than once.
Examples are:
+ 3 + 2 = + 1 + 2
− 9 = + 3 − 3
− 5 + 4 = + 1 − 1 + 2 − 2
In the partial fraction decomposition, each factor becomes the denominator of a smaller rational expression,
and the numerators will be constants. In these examples I’m letting the top be 1. Use the methods I showed
you in class to determine the values of the numerators A, B, etc.:
1
=
+
+ 3 + 2 + 2 + 1
1
=
+
+
− 9 + 3 − 3
1
=
+
+
+
− 5 + 4 + 1 − 1 + 2 − 2
Each numerator gets a single letter representing the unknown value to be determined. The integrals for each
follow. Please check my math for accuracy.
1
1
1
=
−
+
= − ln| + 2| + ln | − 1| + + 3 + 2
+2 +1
1
−1⁄9 1⁄18 1⁄18
1
1
1
= $
+
+
' = − ln|| + ln| + 3| + ln| − 3| + − 9
+3 −3
9
18
18
)
You determine ( * +, - . .
Case II: The polynomial Q (x ) factors into irreducible quadratics.
Irreducible quadratics are those that cannot be factored over the real numbers—equivalently, they have
complex solutions when the quadratic formula is applied. Here are some easy examples of irreducible
quadratics:
+ 1, + 7, + 3 + 5,12.
What about something like + 3 + 5? Rather than laboriously work out the quadratic formula on it, here’s
a fast way to show if this is irreducible: complete the square. If you get something of the form ± ℎ + 5,
where there is a + sign in front of the k, it is irreducible. If there was a negative sign in front of the k, it is
factorable (it may have radical solutions). So, complete the square on + 3 + 5, we get 6 + 7 +
Thus, it is irreducible.
))
.
The numerator of a rational expression with an irreducible quadratic denominator can be linear, so we write
+ , and so forth.
Example:
)
+1
+1
+ =
= + + 8 + 8 + 8
)
The values for A, B and C are = 8, = − 8 and = 1. Therefore, we have:
+1
1⁄8 −1⁄8 + 1
= $
+
' + 8
+ 8
1⁄8
−1⁄8 1
=
+ + +8
+8
1
1
1
= ln|| − ln| + 8| +
tan+) + .
8
16
√8
√8
Case III: Repeated factors (multiplicities).
In the case of repeated linear factors, decompose into smaller rational expressions writing the factor to the 1st
power, then again to the 2nd, etc, to the nth power:
1
1
= = + +
−
+ 1 +1
In the case of repeated irreducible quadratics, write them as above, to the (quantity) 1st power, then 2nd, etc., to
the nth power:
1
+ + >
? + @
=
+ + + − 2 + 1
+ 1
− 2 + 1 + 1
A clever way to know how far out to go is to go out to the nth letter of the alphabet, where n is the degree of
the denominator. In the above example, since the denominator is a 7th-degree polynomial, we know we’ll
)
need to go as far as “G”. For example: +- - .,* has a 10th-power polynomial in the denominator. The 10th
letter of the alphabet is J, so we need to ensure our partial fraction decomposition goes out to include a J in the
numerator. Look closely and you’ll see a linear factor repeated twice, and an irreducible quadratic factor
repeated four times. What is the correct expansion?
Caution: Students sometimes get confused between an irreducible quadratic expression and a repeated linear
)
term repeated. For example, some students would see +- as an irreducible quadratic, which is incorrect. It
is a linear factor, repeated twice. Look carefully.
Case IV: numerator is of degree ≥ than that of the bottom (a so-called improper rational expression). An
example of an improper rational expression is:
+ 3 − 1
+ 5 + 4
We need to long divide the bottom into the top:
The remainder is 24x + 19; therefore, the complete division is:
+ 3 − 1
24 + 19
=−5+ + 5 + 4
+ 5 + 4
To integrate, the x – 5 use regular power rule, and you’d need to use partial fractions on the fractional part.
What do you get?
Answers to questions.
−
1
1
1
1
1
= ln| + 1| − ln| − 1| − ln | + 2| + ln| − 2| + 6
6
12
12
− 5 + 4
1
2 +
5
=
+ > + ?
@ + A
B + C
+
+ + + + + 5
+ 5
− 2 − 2
+ 5 + 5
+ 3 − 1
24 + 19
1
77
5
= − 5 + = − 5 + ln| + 4| − ln| + 1| + + 5 + 4
+ 5 + 4
2
3
3