Partial fractions with simple linear factors
... is a method for computing integrals of the form
Z
Z
P(x)
f (x) dx =
dx.
Q(x)
Here
P and Q are polynomials,
deg(P) < deg(Q), and
Q can be factorized as
Q(x) = (x − a1 ) · · · (x − an ),
where a1 , · · · an are distinct real numbers.
Math 105 (Section 204)
Partial fractions
2011W T2
1/6
Description of the method
Step 1. Obtain a partial fractions expansion for f (x): Find constants
c1 , · · · , cn such that
(∗)
c1
c2
cn
P(x)
=
+
+ ··· +
.
Q(x)
x − a1 x − a2
x − an
Two methods for solving for the constants c1 , · · · , cn :
I
I
equating coefficients of x from both sides, or
choosing values of x = a1 , · · · , an
Step 2. The equation (∗) can be integrated:
Z
n
X
P(x)
cj ln |x − aj | + C .
dx =
Q(x)
j=1
Math 105 (Section 204)
Partial fractions
2011W T2
2/6
Example
Find the partial fractions expansion for
f (x) =
x3
x +1
− 3x 2 + 2x
Step 1. Factorize the denominator:
x 3 − 3x 2 + 2x = x(x 2 − 3x + 2)
= x(x − 1)(x − 2).
Math 105 (Section 204)
Partial fractions
2011W T2
3/6
Example
Find the partial fractions expansion for
f (x) =
x3
x +1
− 3x 2 + 2x
Step 1. Factorize the denominator:
x 3 − 3x 2 + 2x = x(x 2 − 3x + 2)
= x(x − 1)(x − 2).
Step 2. Use the factorization to write down the partial fractions form:
x +1
A
B
C
= +
+
,
x(x − 1)(x − 2)
x
x −1 x −2
where A, B, C are constants to be determined.
Math 105 (Section 204)
Partial fractions
2011W T2
3/6
Example (ctd)
Step 3. Multiply both sides of the equation by x(x − 1)(x − 2) to get
x + 1 = A(x − 1)(x − 2) + Bx(x − 2) + Cx(x − 1).
Set x = 0, then
1 = 2A
so
1
A= .
2
2 = −B
so
B = −2.
3 = 2C
so
3
C= .
2
Set x = 1, then
Set x = 2, then
Thus
Answer
x3
Math 105 (Section 204)
x
1
2
3
=
−
+
.
2
− 3x + 2x
2x
x − 1 2(x − 2)
Partial fractions
2011W T2
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Integration using partial fractions
Find the value of the definite integral
Z 1
0
dt
.
2 + e −t
A. ln 3
B. arctan(e)
C.
1
2
ln
1+2e
3
D. 0.5
Math 105 (Section 204)
Partial fractions
2011W T2
5/6
Integration P(x)/Q(x) when deg(P) ≥ deg(Q)
Find the indefinite integral of
Z
A. ln
h
(x−2)14
|x−1|
B. 3x + ln
h
i
+C
i
14
(x−2)
|x−1|
3x 2 + 4x − 6
dx
x 2 − 3x + 2
+C
C. ln(|x − 1|(x − 2)14 + C
i
h
14
+C
D. ln 3x (x−2)
|x−1|
Math 105 (Section 204)
Partial fractions
2011W T2
6/6