Transcript partial fractions

8.5 day 1 Partial Fractions
The Empire Builder, 1957
Greg Kelly, Hanford High School, Richland, Washington
Partial Fractions
Motivation
The method of partial fractions is a procedure for
decomposing a rational function into simpler rational
functions to which you can apply the basic integration
formulas.
To see the benefit of the method of partial fractions,
consider the integral
Complete the square on the integrand.
5
x
2
2
5 1

x


  
2 2

2

1
2
sec  2x  5
tan   2 x  5x  6
1
x   sec   5 
2
1
2
tan   x  5 x  6
2
1
2
2
tan   x  5 x  6
4
1
dx  sec  tan 
2
2

To evaluate this integral without
partial fractions, you can complete
the square and use trigonometric
substitution to obtain
Partial Fractions
Now, suppose you had observed that
Then you could evaluate the integral easily, as follows.
This method is clearly preferable to trigonometric substitution.
However, its use depends on the ability to factor the denominator,
x2 – 5x + 6, and to find the partial fractions
Linear Factors
1
Write the partial fraction decomposition for
Solution:
Because x2 – 5x + 6 = (x – 3)(x – 2), you should include one
partial fraction for each factor and write
where A and B are to be determined.
Multiplying this equation by the least common denominator
(x – 3)(x – 2) yields the basic equation
1 = A(x – 2) + B(x – 3).
Basic equation.
Because this equation is to be true for all x, you can substitute
any convenient values for x to obtain equations in A and B.
The most convenient values are the ones that make particular
factors equal to 0.
To solve for A, let x = 3 and obtain
1 = A(3 – 2) + B(3 – 3)
1 = A(1) + B(0)
A=1
Let x = 3 in basic equation.
cont’d
To solve for B, let x = 2 and obtain
1 = A(2 – 2) + B(2 – 3)
1 = A(0) + B(–1)
B = –1
So, the decomposition is
as previously shown.
Let x = 2 in basic equation
2
5x  3
 x 2  2 x  3 dx
This would be a lot easier if we could
re-write it as two separate terms.
5x  3
A
B


 x  3 x  1 x  3 x  1
Non-repeating linear factors.

2
5x  3
 x 2  2 x  3 dx
The short-cut for this type of problem is
called the Heaviside Method, after
English engineer Oliver Heaviside.
5x  3
A
B


 x  3 x  1 x  3 x  1
5x  3  A  x  1  B  x  3
Multiply by the common
denominator.
Let x = - 1
8  A  0  B  4
2B
12  A   4   B  0
3 A
Let x = 3

2
5x  3
 x 2  2 x  3 dx
The short-cut for this type of problem is
called the Heaviside Method, after
English engineer Oliver Heaviside.
5x  3
A
B


 x  3 x  1 x  3 x  1
5x  3  A  x  1  B  x  3
8  A  0  B  4
2B
12  A   4   B  0
3 A
3
2
 x  3  x  1 dx
3ln x  3  2ln x  1  C

Good News!
The AP Exam only requires non-repeating linear factors!
The more complicated methods of partial fractions are
good to know, and you might see them in college, but they
will not be on the AP exam or on my exam.

3
6x  7
 x  2
2
A
B


2
x  2  x  2
6 x  7  A  x  2  B
6x  7  Ax  2 A  B
6x  Ax
7  2A  B
6 A
7  26  B
7  12  B
5  B
Repeated roots: we must
use two terms for partial
fractions.
6
5

2
x  2  x  2
6x  7
  x  2
2
dx
6
5


dx
2
x  2  x  2
5
 6 ln x  2 
C
x2

4
2 x3  4 x 2  x  3
 x2  2 x  3 dx
If the degree of the numerator is
higher than the degree of the
denominator, use long division first.
2x
x 2  2 x  3 2 x3  4 x 2  x  3
2 x3  4 x 2  6 x
5x  3
5x  3
 2 x  x 2  2 x  3 dx
5x  3
  2x 
dx
 x  3 x  1
(from example 2)
3
2
  2x 

dx  x2  3ln  x  3  2ln  x  1  C
 x  3  x  1

5
You Try:

5 x  20 x  6
2

x  x  1
2
5x  20x  6
dx
3
2
x  2x  x
2
A
B
C
 

2
x x  1 x  1
5 x  20 x  6  Ax  1  Bx x  1  Cx
2
2
A  6, B  1, C  9

6
1
9
5x  20x  6

dx
dx   
2
3
2
x x  1 ( x  1)
x  2x  x
2
x  1

 6ln x  ln x  1  9
1
1

x6
9
 ln

C
x 1 x 1
C
Homework
• Section 8.5 Day 1: pg. 559, 7-19 odd, 29,30,43
• Day 2: MMM BC pgs. 115-116
Quadratic Factors
6
Distinct Linear and Quadratic Factors
Find
Solution:
Because (x2 – x)(x2 + 4) = x(x – 1)(x2 + 4) you should include one
partial fraction for each factor and write
Multiplying by the least common denominator x(x – 1)(x2 + 4)
yields the basic equation:
2x3 – 4x – 8 = A(x – 1)(x2 + 4) + Bx(x2 + 4) + (Cx + D)(x)(x – 1)
6
To solve for A, let x = 0 and obtain
–8 = A(–1)(4) + 0 + 0
2=A
To solve for B, let x = 1 and obtain
–10 = 0 + B(5) + 0
–2 = B
At this point, C and D are yet to be determined.
You can find these remaining constants by choosing two other
values for x and solving the resulting system of linear equations.
6
If x = –1, then, using A = 2 and B = –2, you can write
–6 = (2)(–2)(5) + (–2)(–1)(5) + (–C + D)(–1)(–2)
2 = –C + D
If x = 2, you have
0 = (2)(1)(8) + (–2)(2)(8) + (2C + D)(2)(1)
8 = 2C + D
Solving the linear system by subtracting the first equation from
the second
–C + D = 2
2C + D = 8
yields C = 2
6
Consequently, D = 4, and it follows that
Homework
• Section 8.5 Day 1: pg. 559, 7-21 odd, 29,30
• Day 2: MMM BC pgs. 115-116