Partial Fraction Decomposition

Integrating rational functions.

Integrate the following function…..

∫ 1𝑥−3 −

1𝑥−2 𝑑𝑥

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

1𝑥2−5 𝑥+6

=𝐴 (𝑥−2 )+𝐵 (𝑥−3 )

(𝑥−3 ) (𝑥+2 )

¿

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.

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

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

∫ 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

∫ 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

rules• Remember is just

• If , then

• When thinking about your natural log rules, use and to guide you.

Which of the following are true?

or

or

Which of the following are true?

or

or

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

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)

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

3 ln (𝑥+2 )+ ln (𝑥−2 )+2 ln (𝑥−1 )+𝐶

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…….

Long “Polynomial” Division Review

𝒙𝟐−𝟗 𝒙−𝟏𝟎𝒙+𝟏

𝒙+𝟏√𝒙𝟐−𝟗 𝒙−𝟏𝟎

𝑥2+9 𝑥+14𝑥+7

𝒙+𝟕√𝒙𝟐+𝟗 𝒙+𝟏𝟒

3𝑥3−5 𝑥2+10𝑥−33 𝑥+1

𝟑 𝒙+𝟏√𝟑 𝒙𝟑−𝟓 𝒙𝟐+𝟏𝟎𝒙−𝟑

∫ 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

Since no we must𝒙Put .𝟎𝒙2𝑥3−9 𝑥2+15

2𝑥−5

∫ 2 𝑥3−9𝑥2+152𝑥−5 𝑑𝑥

∫(𝑥2−2𝑥−5− 102𝑥−5 )𝑑𝑥

4 𝑥4+3𝑥3+2𝑥+1𝑥2+𝑥+2

Since no we mustPut

End of Lesson

Homework:Partial Fraction Decomp. Worksheet (14, 15)

Orange Book Section 6.5 P. 369 (5, 7, 8, 15-21 odd)

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.