Partial Fractions. Understand the concept of partial fraction decomposition. Use partial fraction...
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![Page 1: Partial Fractions. Understand the concept of partial fraction decomposition. Use partial fraction decomposition with linear factors to integrate rational.](https://reader036.fdocuments.net/reader036/viewer/2022062303/56649f2f5503460f94c4932b/html5/thumbnails/1.jpg)
Partial Fractions
![Page 2: Partial Fractions. Understand the concept of partial fraction decomposition. Use partial fraction decomposition with linear factors to integrate rational.](https://reader036.fdocuments.net/reader036/viewer/2022062303/56649f2f5503460f94c4932b/html5/thumbnails/2.jpg)
Understand the concept of partial fraction decomposition.
Use partial fraction decomposition with linear factors to integrate rational functions.
Use partial fraction decomposition with quadratic factors to integrate rational functions.
Objectives
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Partial Fractions
![Page 4: Partial Fractions. Understand the concept of partial fraction decomposition. Use partial fraction decomposition with linear factors to integrate rational.](https://reader036.fdocuments.net/reader036/viewer/2022062303/56649f2f5503460f94c4932b/html5/thumbnails/4.jpg)
Partial Fractions
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
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Partial Fractions
Now, suppose you had observed that
Then you could evaluate the integral easily, as follows.
However, its use depends on the ability to factor the denominator, x2 – 5x + 6, and to find the partial fractions
Partial fraction decomposition
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Partial Fractions
![Page 7: Partial Fractions. Understand the concept of partial fraction decomposition. Use partial fraction decomposition with linear factors to integrate rational.](https://reader036.fdocuments.net/reader036/viewer/2022062303/56649f2f5503460f94c4932b/html5/thumbnails/7.jpg)
Linear Factors
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Example 1 – Distinct Linear Factors
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.
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Example 1 – SolutionTo solve for A, let x = 3 and obtain
0x+1 = Ax+Bx-2A – 3B
Equating x0 terms
1 = -2A -3B (1)
and Equating x1 terms
0 = A+B
so -A=B (2)
Substitution of (2) into (1) yields
1=2B-3B
1=-1B
thus -1=B (3)
and A=1 (4)
cont’d
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Example 1 – Solution
So, the decomposition is
Restating the original equation
and substituting A=1, B=-1 as shown at the beginning of this section, we get
cont’d
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Quadratic Factors
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Example 3 – 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)
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Example 3 – SolutionEquate X3 terms
2x3=Ax3+Bx3+Cx3
so 2=A+B+C (1)
Equate X2 terms
0=-AX2-Cx2+Dx2
so 0=-A-C+D (2)
Equate X terms
-4X=4AX+4BX-DX
so -4=4A+4B-D (3)
Equate X0 terms
-8=-4A (4)
cont’d
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Example 3 – SolutionSolving for A
-8=-4A
A=2 (5)
Substitution of (5) into (3)
-4=4A+4B-D
-4=8+4B-D
-12=4B-D (6)
and substitution of (5) into (1)
2=A+B+C
2=2+B+C
0=B+C
B=-C (7)
cont’d
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Example 3 – SolutionSubstitution of (5) into (2)
-0=-2-C+D
2=-C+D and substitution of (7) yields
2=B+D
2-B=D (8)
Now substitution of (8) and (6)
-12=4B-D
-12=4B-(2-B)
-12=4B-2+B
-12=5B-2
-10=5B
so B=-2 (9)
cont’d
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Example 3 – Solution
Substitution of (9) into (7)
B=-C
So C=2 (10)
Solving (8)
2-B=D
2-(-2)=D
So D=-4
restating the problem
cont’d
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Example 3 – Solution
Substituting A=2, B=-2, C=2, and D = 4, it follows that
cont’d