Download Partial Fraction Decomposition

Partial Fraction Decomposition
In order to add two rational functions you first find a common denominator and then
combine and simplify the numerator.
x2
1
x3 +1 + x2 +1
=
=
x2 (x2 +1)+1(x3 +1)
(x3 +1)(x2 +1)
4
x +x3 +x2 +1
x5 +x3 +x2 +1
The idea behind partial fraction decomposition is to undo this process. We do three mayor
types of problems in this class:
1
1. Non-repeated linear factors, e.g. (x−1)(x+2)
.
1
2. Repeated linear factors, e.g. (x−1)(x+2)
2.
1
3. Irreducible quadratic factors, e.g. (x−1)(x
2 +1) .
Below is an example of each type of decomposition. Remember, when you go to do other
problems, that the degree of the numerator must be less than the degree of the denominator.
Example 1
Z
To find the integral
1
(x−1)(x+2) dx we start by decomposing the fraction.
1
(x−1)(x+2)
A
B
= x−1
+ x+2
=
A(x+2)+B(x−1)
(x−1)(x+2)
Therefore,
A(x + 2) + B(x − 1) = 1.
Now, if x = 1 then we get
and if x = −2 then we get
3A = 1 and A = 13 ,
−3B = 1 and B = −1
3 .
So,
Z
1
(x−1)(x+2) dx
=
1
3
Z
1
1
x−1 dx − 3
Z
1
x+2 dx
= 13 ln(|x − 1|) − 13 ln(|x + 2|) + c.
Example 2
Find the integral
Z
x2
(x−1)(x−2)2 . Again we start by decomposing the fraction.
x2
(x−1)(x−2)2
A
B
C
= x−1
+ x−2
+ (x−2)
2
=
A(x−2)2 +B(x−1)(x−2)+C(x−1)
(x−1)(x−2)2
Therefore,
A(x − 2)2 + B(x − 1)(x − 2) + C(x − 1) = x2 .
Now, if x = 2 then we get
C = 4,
if x = 1 then
A = 1,
and if x = 0 then
4 + 2B − 4 = 0 and B = 0.
So,
Z
x2
(x−1)(x−2)2
Z
=
1
x−1 dx + 4
Z
1
(x−2)2 dx
4
= ln(|x − 1|) − x−2
+ c.
Example 3
Z
Find the integral
x
(x−1)(x2 +1) dx. First we decompose the fraction.
x
(x−1)(x2 +1)
A
= x−1
+ Bx+C
x2 +1
=
A(x2 +1)+(Bx+C)(x−1)
(x−1)(x2 +1)
Therefore,
A(x2 + 1) + (Bx + C)(x − 1) = x.
Now, if x = 1 then
2A = 1 and A = 12 ,
if x = 0 then
1
1
2 − C = 0 and C = 2 ,
and if x = 2 then
5
1
−1
2 + 2B + 2 = 2 and B = 2 .
So,
Z
x
(x−1)(x2 +1) dx
=
1
2
=
1
2
Z
Z
1
1
x−1 dx + 2
1
1
x−1 dx + 2
Z
Z
1−x
x2 +1 dx
1
1
x2 +1 dx − 2
Z
x
x2 +1 dx
= 21 ln(|x − 1|) + 12 arctan(x) − 14 ln(|x2 + 1|) + c
Trigonometric Substitution
Trigonometric Substitution, trig sub, uses trig identities to simplify integrals and make them
easier to manage. In particular you use:
1. a2 − a2 sin2 (θ) = a2 cos2 (θ)
2. a2 + a2 tan2 (θ) = a2 sec2 (θ)
3. a2 sec2 (θ) − a2 = a2 tan2 (θ)
To solve problems using trig sub you use one of three possible substitutions:
1. if you have an integral with a a2 − u2 then use u = a sin(θ),
2. if you have an integral with a u2 + a2 then use u = a tan(θ),
3. if you have an integral with a u2 − a2 then use u = a sec(θ).
Z
√ 1
Find the integral of
dx. Since we have something that looks like a2 − u2 we will
2
1−x
use the substitution u = a sin(θ).
x = sin(θ)
dx = cos(θ) dθ
1 − x2 = 1 − sin2 (θ) = cos2 (θ)
And, we integrate,
Z
√ 1
dx
1−x2
Z
=
Z
=
Z
=
√cos(θ)
dθ
2
cos (θ)
cos(θ)
cos(θ) dθ
dθ
= θ+c
= arcsin(θ) + c
Question: The function sin(θ) is only defined between 1 and −1, however x may
be defined everywhere, so how can we make the substitution x = sin(θ)?