9.2 – Adding & Subtracting Rational Expressions. Remember adding & subtracting fractions?
Integrating Rational Functions by Partial Fractions Objective: To make a difficult/impossible...
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Integrating Rational Functions by Partial Fractions Objective: To make a difficult/impossible integration problem easier.
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Transcript of Integrating Rational Functions by Partial Fractions Objective: To make a difficult/impossible...
Integrating Rational Functions by Partial Fractions
Objective: To make a difficult/impossible integration
problem easier.
Partial Fractions
• In algebra, you learn to combine two or more fractions into a single fraction by finding a common denominator. For example
43
105
)1)(4(
)4(3)1(2
1
3
4
22
xx
x
xx
xx
xx
Partial Fractions
• However, for the purposes of integration, the left side of this equation is preferable to the right side since each term is easy to integrate.
Cxxdxx
dxx
|1|ln3|4|ln2
1
3
4
2
43
105
)1)(4(
)4(3)1(2
1
3
4
22
xx
x
xx
xx
xx
Partial Fractions
• We need a method to take and make it
• This method is called Partial Fractions. This method only works for proper rational fractions, meaning that the degree of the numerator is less than the degree of the denominator. This is how it works.
43
1052
xx
x
.1
3
4
2
xx
Partial Fractions
• Factor the denominator completely.
)1)(4(
105
43
1052
xx
x
xx
x
Partial Fractions
• Factor the denominator completely.
• Assign a variable as the numerator to each term of the denominator and set it equal to the original.
)1)(4(
105
43
1052
xx
x
xx
x
14)1)(4(
105
x
B
x
A
xx
x
Partial Fractions
• Multiply by the common denominator.
)4()1(105 xBxAx14)1)(4(
105
x
B
x
A
xx
x
Partial Fractions
• Multiply by the common denominator.
• Solve for A and B. • To solve for A, let x = 4, which gives us
• To solve for B, let x = -1, which gives us
A
A
2
510
)4()1(105 xBxAx14)1)(4(
105
x
B
x
A
xx
x
B
B
3
515
Cxxdxx
dxx
|1|ln3|4|ln2
1
3
4
2
Example 1
• Evaluate 22 xx
dx
Example 1
• Evaluate 22 xx
dx
)1)(2(22 xx
dx
xx
dx
Example 1
• Evaluate 22 xx
dx
)1)(2(22 xx
dx
xx
dx12)1)(2(
1
x
B
x
A
xx
Example 1
• Evaluate 22 xx
dx
)1)(2(22 xx
dx
xx
dx12)1)(2(
1
x
B
x
A
xx
)2()1(1 xBxA
Example 1
• Evaluate 22 xx
dx
)1)(2(22 xx
dx
xx
dx12)1)(2(
1
x
B
x
A
xx
)2()1(1 xBxA
B
B
x
3/1
31
1
A
A
x
3/1
31
2
Example 1
• Evaluate 22 xx
dx
|2|ln3
1|1|ln
3
1
2
1
3
1
1
1
3
1
xxxx
Cx
x
2
1ln3
1
Linear Factors
• Linear Factor Rule.• For each factor of the form , the partial
fractions decomposition contains the following sum of m partial fractions
where A1, A2, …Am are constants to be determined. In the case where m = 1, only the first term appears.
mm
bax
A
bax
A
bax
A
)(...
)( 221
mbax )(
Example 2
• Evaluate dxxx
x
23 2
42
Example 2
• Evaluate dxxx
x
23 2
42
)2(
422 xx
x
Example 2
• Evaluate dxxx
x
23 2
42
)2(
422 xx
x
2)2(
422
x
C
x
B
x
A
xxx
x
Example 2
• Evaluate dxxx
x
23 2
42
)2(
422 xx
x
2)2(
422
x
C
x
B
x
A
xxx
x
2)2()2(42 CxxBxAxx
Example 2
• Evaluate dxxx
x
23 2
42
)2(
422 xx
x
2)2(
422
x
C
x
B
x
A
xxx
x
2)2()2(42 CxxBxAxx
B
B
x
2
24
0
C
c
x
2
48
2
Example 2
• Evaluate
• Since there is no way to isolate A, we need to solve with a different method. Let x = 1, substitute our values for B and C and solve for A.
dxxx
x
23 2
42
2)2()2(42 CxxBxAxx B
B
x
2
24
0
C
c
x
2
48
2
Example 2
• Evaluate
• Since there is no way to isolate A, we need to solve with a different method. Let x = 1, substitute our values for B and C and solve for A.
dxxx
x
23 2
42
2)2()2(42 CxxBxAxx B
B
x
2
24
0
C
c
x
2
48
2
2)1(2)1)(2()1(6 A
A
A
2
46
Example 2
• Evaluate dxxx
x
23 2
42
B
B
x
2
24
0
C
c
x
2
48
2
2222
2 x
dx
x
dx
x
dx
A
A
2
46
Cxx
x |2|ln22
||ln2
Cx
x
x
|
2|ln2
2
Example 2-WRONG
• Evaluate dxxx
x
23 2
42
)2(
422 xx
x
2)2(
42
x
C
x
B
x
A
xxx
x
2)2()2(42 CxxBxxAxx
Example 2-WRONG
• Evaluate
• What next? This doesn’t work!
dxxx
x
23 2
42
)2(
422 xx
x
2)2(
42
x
C
x
B
x
A
xxx
x
2)2()2(42 CxxBxxAxx
C
c
x
2
48
2
Example 2-WRONG
• Evaluate
• The denominator on the right is• The denominator on the left is• They are not the same!!
dxxx
x
23 2
42
)2(
422 xx
x
2)2(
42
x
C
x
B
x
A
xxx
x
)2( xx)2(2 xx
Quadratic Factors
• Quadratic Factor Rule• For each factor of the form , the partial
fraction decomposition contains the following sum of m partial fractions:
where A1, A2,…Am, B1, B2,…Bm are constants to be determined. In the case where m = 1, only the first term appears.
mcbxax )( 2
mmm
cbxax
BxA
cbxax
BxA
cbxax
BxA
)(...
)( 22222
211
Example 3
• Evaluate
dxxxx
xx
133
223
2
Example 3
• Evaluate • Factor by grouping
dxxxx
xx
133
223
2
113)1)(13(
222
2
x
CBx
x
A
xx
xx
)1)(13()13(1)13( 22 xxxxx
Example 3
• Evaluate • Factor by grouping
dxxxx
xx
133
223
2
113)1)(13(
222
2
x
CBx
x
A
xx
xx
)1)(13()13(1)13( 22 xxxxx
)13)(()1(2 22 xCBxxAxx
Example 3
• Evaluate • Multiply the right side of the equation and group the
terms based on powers of x.
dxxxx
xx
133
223
2
)13)(()1(2 22 xCBxxAxx
CCxBxBxAAxxx 332 222
)()3()3(2 22 CAxBCxBAxx
Example 3
• Evaluate • Set the coefficients from the right side of the equation
equal to the ones on the left side.
dxxxx
xx
133
223
2
)()3()3(2 22 CAxBCxBAxx
BA 31
BC 31
CA 2
Example 3
• Evaluate • Take Eq 1 – Eq 3
dxxxx
xx
133
223
2
BA 31
BC 31
CA 2
BC 33 BA 31 )2( CA
Example 3
• Evaluate • Take Eq 1 – Eq 3
• Take new Eq + 3Eq 2
dxxxx
xx
133
223
2
BA 31
BC 31
CA 2
BC 33 C
C
5/3
106
BA 31 )2( CA
BC 393
BC 33
Example 3
• Evaluate • Take Eq 1 – Eq 3
• Take new Eq + 3Eq 2
dxxxx
xx
133
223
2
BA 31
BC 31
CA 2
BC 33
C
C
5/3
106
BA 31 )2( CA
BC 393
BC 33
A 5/7
B5/4
Example 3
• Evaluate
dxxxx
xx
133
223
2
dxx
xdx
xdx
xx
xx
1
5/3)5/4(
13
5/7
)1)(13(
222
2
dxx
dxx
xdx
xdx
xx
xx
1
5/3
1
)5/4(
13
5/7
)1)(13(
2222
2
Example 3
• Evaluate
dxxxx
xx
133
223
2
dxx
xdx
xdx
xx
xx
1
5/3)5/4(
13
5/7
)1)(13(
222
2
dxx
dxx
xdx
xdx
xx
xx
1
5/3
1
)5/4(
13
5/7
)1)(13(
2222
2
Cxxx 12 tan5
3)1ln(
5
2|13|ln
15
7
Example 4
• Evaluate
dxxx
xxxx
22
234
)3)(2(
9201643
Example 4
• Evaluate
dxxx
xxxx
22
234
)3)(2(
9201643
22222
234
)3(32)3)(2(
9201643
x
EDx
x
CBx
x
A
xx
xxxx
Homework
• Section 7.5• Page 521• 1-19 odd
