Partial Fraction Decomposition
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Transcript of Partial Fraction Decomposition
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Partial Fraction Decomposition
Integrating rational functions.
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Integrate the following function…..
∫ 1𝑥−3 −
1𝑥−2 𝑑𝑥
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Now what if we disguise this equation a bit by combining……
∫ 1𝑥−3 −
1𝑥−2 𝑑𝑥=∫ 1 (𝑥−2 )−1 (𝑥−3 )
(𝑥−3)(𝑥−2)𝑑𝑥∫ 𝑥−2−𝑥+3
𝑥2−5 𝑥+6𝑑𝑥=∫ 1
𝑥2−5𝑥+6𝑑𝑥
If we are given this equation initially:
We are going to have to do some expansion in orderto put it into a form that is easy to integrate.This process is called PARTIAL FRACTION DECOMPOSITION.
∫ 1𝑥2−5 𝑥+6
𝑑𝑥
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1𝑥2−5 𝑥+6
=𝐴 (𝑥−2 )+𝐵 (𝑥−3 )
(𝑥−3 ) (𝑥+2 )
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¿
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The method of Partial Fraction Decomposition ALWAYS works when you are integrating a rational function.
Rational Function = Ratio of polynomials
You will decompose/expand the rational function so it can be easily integrated.
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4𝑥2−1
=𝐴
𝑥+1 +𝐵
𝑥−1
4=𝐴 (𝑥−1 )+𝐵(𝑥+1)Let 𝐴=−2Let 𝐵=2
∫( −𝟐𝒙+𝟏+ 𝟐𝒙−𝟏 )𝒅𝒙
∫ 𝟏𝒙𝟐−𝟏
𝒅𝒙=−𝟐𝐥𝐧 (𝒙+𝟏 )+𝟐𝐥𝐧 (𝒙−𝟏 )+𝑪
∫ 4𝑥2−1
𝑑𝑥
2𝐵=4 −2 𝐴=4
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74 𝑥2−9
=𝐴
2 𝑥+3 +𝐵
2 𝑥−3
7=𝐴 (2𝑥−3 )+𝐵(2𝑥+3)
Let 7=6𝐵𝐵=76
Let 7=−6 𝐴𝐴=− 76
14 𝑥2−9
=−7 /62 𝑥+3 +
7 /62 𝑥−3
− 712 ln(2 𝑥+3 )+ 712 ln
(2 𝑥−3 )+𝐶
∫ 74 𝑥2−9
𝑑𝑥
∫ −7 /62 𝑥+3 +
7 /62 𝑥−3 𝑑𝑥
*Integrate using U-Substitution
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∫ 5 𝑥−15𝑥2−𝑥−12
𝑑𝑥5 𝑥−15𝑥2−𝑥−12
=𝐴
𝑥−4 +𝐵𝑥+3
5 𝑥−15=𝐴 (𝑥+3 )+𝐵(𝑥−4)Let 5 (−3 )−15=𝐵(−7)Let 5 (4 )−15=𝐴(7)
𝐴=57 𝐵=
307
∫ 5 𝑥−15𝑥2−𝑥−12
𝑑𝑥=∫ 5 /7𝑥−4 +
30/7𝑥+3 𝑑𝑥
𝟓𝟕 𝐥𝐧 (𝒙−𝟒 )+𝟑𝟎𝟕 𝐥𝐧 (𝒙+𝟑 )+𝑪
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∫ 3 𝑥−8𝑥2−4 𝑥−5
𝑑𝑥
3𝑥−8𝑥2−4 𝑥−5
=𝐴
𝑥−5+𝐵𝑥+1
3 𝑥−8=𝐴 (𝑥+1 )+𝐵 (𝑥−5 )
Let
3 (−1 )−8=𝐵 (−6 ) ∴𝐵=116 3 (5 )−8=𝐴 (6 )∴ 𝐴=
76
∫ 3 𝑥−8𝑥2−4 𝑥−5
𝑑𝑥=∫76
𝑥−5 +
116𝑥+1
𝑑𝑥
76 ln
(𝑥−5 )+ 116 ln(𝑥+1 )+𝐶
Let
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rules• Remember is just
• If , then
• When thinking about your natural log rules, use and to guide you.
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Which of the following are true?
or
or
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Which of the following are true?
or
or
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HomeworkSection 8.5 P. 559 (7-11)
Section 5.6P. 378 (45, 59)
Section 5.7P. 385 (3, 4, 7)
Partial Fraction Decomposition
Derivating Arctangent
Integrating Arctangent
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𝑑𝑦𝑑𝑥=
6 𝑥2−8 𝑥−4(𝑥2−4 ) (𝑥−1 )
6 𝑥2−8 𝑥−4(𝑥2−4 ) (𝑥−1 )
=𝐴
𝑥+2+
𝐵𝑥−2 +
𝐶𝑥−1
(𝑥2−4 ) (𝑥−1 )=(𝑥+2)(𝑥−2)(𝑥−1)
𝟔 𝒙𝟐−𝟖 𝒙−𝟒=𝑨 (𝒙−𝟐 ) (𝒙−𝟏 )+𝑩 (𝒙+𝟐 ) (𝒙−𝟏 )+𝑪(𝒙+𝟐)(𝒙 −𝟐)
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Let 6 (−2 )2−8 (−2 )−4=𝐴(−4)(−3)36=12 𝐴∴ 𝐴=3
Let 6 (2 )2−8 (2 )−4=𝐵(4)(1)
4=4𝐵∴𝐵=1
Let 6 (1 )2−8−4=𝐶 (3)(−1)
−6=−3𝐶∴𝐶=2
6 𝑥2−8 𝑥−4(𝑥2−4 ) (𝑥−1 )
=3
𝑥+2+1
𝑥−2 +2
𝑥−1
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3 ln (𝑥+2 )+ ln (𝑥−2 )+2 ln (𝑥−1 )+𝐶
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In all of our examples thus far, the degree of the numerator has been less than the degree of the denominator.
If it is the case that the degree of the numerator is greater than or equal to the degree of the denominator, you must reduce using “polynomial” long division.
The next few slides will help you to review this technique…….
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Long “Polynomial” Division Review
𝒙𝟐−𝟗 𝒙−𝟏𝟎𝒙+𝟏
𝒙+𝟏√𝒙𝟐−𝟗 𝒙−𝟏𝟎
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𝑥2+9 𝑥+14𝑥+7
𝒙+𝟕√𝒙𝟐+𝟗 𝒙+𝟏𝟒
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3𝑥3−5 𝑥2+10𝑥−33 𝑥+1
𝟑 𝒙+𝟏√𝟑 𝒙𝟑−𝟓 𝒙𝟐+𝟏𝟎𝒙−𝟑
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∫ 3 𝑥3−5𝑥2+10 𝑥−33𝑥+1
dx
∫𝑥2−2𝑥+4− 73 𝑥+1 𝑑𝑥
𝟏𝟑 𝒙𝟑−𝒙𝟐+𝟒 𝒙−𝟕𝟑 𝐥𝐧 (𝟑𝒙+𝟏 )+𝑪
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Since no we must𝒙Put .𝟎𝒙2𝑥3−9 𝑥2+15
2𝑥−5
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∫ 2 𝑥3−9𝑥2+152𝑥−5 𝑑𝑥
∫(𝑥2−2𝑥−5− 102𝑥−5 )𝑑𝑥
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4 𝑥4+3𝑥3+2𝑥+1𝑥2+𝑥+2
Since no we mustPut
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End of Lesson
Homework:Partial Fraction Decomp. Worksheet (14, 15)
Orange Book Section 6.5 P. 369 (5, 7, 8, 15-21 odd)
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Steps to Integrating by PFD1. If degree of numerator is greater than or equal
to degree of denominator, then use long division to reduce.
2. Write out or setup the equation as a sum of fractions with unknown numerators.
3. Solve for the unknown numerators.
4. Integrate the resulting equation.