# About Partial Fractions 1. Distinct Linear Factors: If p(x) is a

```About Partial Fractions
1. Distinct Linear Factors: If p(x) is a polynomial with degree less than
n and r1 , r2 , . . . , rn are n distinct numbers, then there are numbers A1 ,
A2 , A3 , . . . , An such that
A1
A2
An
p(x)
=
+
+ &middot;&middot;&middot; +
(x − r1 )(x − r2 ) &middot; &middot; &middot; (x − rn )
x − r1
x − r2
x − rn
2. Repeated Linear Factors: If p(x) is a polynomial with degree less than
n and r is a number, then there are numbers A1 , A2 , A3 , . . . , An such that
A1
A2
A3
An
p(x)
=
+
+
+ &middot;&middot;&middot; +
n
2
3
(x − r)
x − r (x − r)
(x − r)
(x − r)n
3. Repeated Quadratic Factors: If p(x) is a polynomial with degree less
than 2n and ax2 + bx + c is a polynomial which is never zero, then there
are numbers A1 , A2 , A3 , . . . , An , B1 , B2 , B3 , . . . , Bn so that
(ax2
p(x)
A1 x + B1
A2 x + B2
A3 x + B 3
An x + Bn
= 2
+
+
+&middot; &middot; &middot;+
n
2
2
2
3
+ bx + c)
ax + bx + c (ax + bx + c) (ax + bx + c)
(ax2 + bx + c)n
4. We can factor a polynomial into a product of factors each of them is either
linear (i.e., ax + b with a 6= 0), repeated linear (i.e., (ax + b)n with a 6= 0),
quadratic (i.e., ax2 + bx + c with b2 − 4ac &lt; 0) and repeated quadratic
(i.e., (ax2 + bx + c)n with b2 − 4ac &lt; 0). For example,
(x − 3)(x − 4)(x − 5)(x − 8)3 (x2 + 1)(x2 + 4x + 12)5 =
(x − 3)
| {z }
Factors with three as a root
(x − 4)
| {z }
Factors with four as a root
(x − 5)
| {z }
Factors with four as a root
3
(x − 8)
| {z }
Factors with eight as a root
x2 + 1
| {z }
Factors involving x2 + 1
(x2 + 4x + 12)5
|
{z
}
Factors involving x2 + 4x + 12
5. We don’t use this! A decomposition which we do not use:
7x3 + 1
(x − 3)(x − 4)(x − 5)(x − 8)3 (x2 + 1)(x2 + 4x + 12)5
A
B
C
=
+
+
x−3 x−4 x−5
Dx2 + Ex + F
Gx + H
+
+ 2
(x − 8)3
x +1
9
8
7
Ix + Jx + Kx + Lx6 + M x5 + N x4 + P x3 + Qx2 + Rx + S
+
(x2 + 4x + 12)5
We don’t use this because the integrals involving D through F , and
I through Q are still difficult looking. Notice that there are 18 constants
to solve for (A, B, through S, but not O) and the the degree of the
denominator of the left hand side is 18.
6. Partial Fraction Decomposition: Suppose p(x)
q(x) is a proper rational
function. Factor the denominator as much as possible. We can factor the
denominator as product of linear factors, repeated linear factors, quadratic
factors and repeated quadratic factors. For each of these denominators,
write down the partial fraction decomposition and add them. For example,
7x3 + 1
(x − 3)(x − 4)(x − 5)(x − 8)3 (x2 + 1)(x2 + 4x + 12)5
A
B
C
=
+
+
x−3 x−4 x−5
E
D
F
Gx + H
+
+ 2
+
+
2
3
x − 8 (x − 8)
(x − 8)
x +1
Ix + J
Kx + L
Mx + N
+ 2
+
+ 2
x + 4x + 15 (x2 + 4x + 15)2
(x + 4x + 15)3
Px + Q
Rx + S
+ 2
+ 2
(x + 4x + 15)4
(x + 4x + 15)5
Notice that there are 18 constants to solve for (A, B, through S, but not
O) and the the degree of the denominator of the left hand side is 18. Also,
there are 12 things added up here and there are 12 things which are multiplied together to give the denominator.
Remember that you can also use techniques like substitution to compute
the integral of a proper rational function.
7. The values for D, E and F for the two equations above will not be equal.
8. The values of I through S (except O) for the two equations above will not
be equal.
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