Section 8.4aPARTIAL FRACTIONS

A flashback to Section 6.5…We evaluated the following integral:

100100

dPP P

1 1100

dPP P

ln ln 100P P C

This expansion technique is the method of partial fractions.Any rational function can be written as a sum of basicfractions, called partial fractions, using this method.

Method of Partial Fractions (f(x)/g(x) Proper)Notes: The degree of f(x) must be less than the degree of g(x)(if it is not, do long division!). Also, g(x) must be factorable.

x r1. Let be a linear factor of . Suppose is the highest power of that divides . Then, to this factor, assign the sum of the partial fractions:

g x mx r x r

g x m

1 2

2m

m

AA Ax r x r x r

Do this for each distinct linear factor of . g x

Method of Partial Fractions (f(x)/g(x) Proper)Notes: The degree of f(x) must be less than the degree of g(x)(if it is not, do long division!). Also, g(x) must be factorable.

2x px q 2. Let be a quadratic factor of .Suppose is the highest power of thisfactor that divides . Then, to this factor, assignthe sum of the partial fractions:

1 1 2 2

22 2 2

n nn

B x CB x C B x Cx px q x px q x px q

Do this for each distinct quadratic factor of that cannot be factored into linear factors with realcoefficients.

g x

g x 2 nx px q

g xn

Method of Partial Fractions (f(x)/g(x) Proper)Notes: The degree of f(x) must be less than the degree of g(x)(if it is not, do long division!). Also, g(x) must be factorable.

3. Set the original fraction equal to thesum of all these partial fractions. Clear the resultingequation of fractions and arrange the terms indecreasing powers of .

f x g x

x

4. Equate the coefficients of corresponding powersof and solve the resulting equations for theundetermined coefficients.x

The new fractions should be easier to integrate!!!

Guided PracticeUse partial fractions to evaluate the integral.

2

5 32 3x dx

x x

2

5 32 3x

x x

5 31 3x

x x

1 3

A Bx x

5 3 3 1x A x B x

5 3 3x Ax A Bx B

5 3 3x A B x A B

5A B 3 3A B

Solve the system!!!2, 3A B

Guided PracticeUse partial fractions to evaluate the integral.

2

5 32 3x dx

x x

1 3A B dxx x

2 31 3

dxx x

2ln 1 3ln 3x x C

Guided PracticeUse partial fractions to evaluate the integral.

26 72

x dxx

26 72

xx

22 2

A Bx x

6 7 2x A x B 6 7 2x Ax A B

6A

6 7 2x A x A B

2 7A B

Solve the system!!!6, 5A B

Guided PracticeUse partial fractions to evaluate the integral.

26 72

x dxx

22 2

A B dxx x

26 52 2

dxx x

26 5 22

dx x dxx

16ln 2 5 2x x C

Guided PracticeUse partial fractions to evaluate the integral.

3 2

2

2 4 32 3

x x x dxx x

The degree of the numerator is larger than the degree ofthe denominator… we need long division first:

2 3 22 3 2 4 3x x x x x 2x

3 22 4 6x x x 5 3x

2

5 322 3xx dx

x x

Guided PracticeUse partial fractions to evaluate the integral.

3 2

2

2 4 32 3

x x x dxx x 2

5 322 3xx dx

x x

2

5 322 3xxdx dx

x x

2 321 3

xdx dxx x

2 2ln 1 3ln 3x x x C

Guided PracticeUse partial fractions to evaluate the integral.

3 2

22

2 2

1

x x dxx

3 2

22

2 2

1

x x

x

22 21 1

Ax B Cx Dx x

3 2 22 2 1x x Ax B x Cx D 3 2 3 22 2x x Ax Bx Ax B Cx D

3 2 3 22 2x x Ax Bx A C x B D

Guided PracticeUse partial fractions to evaluate the integral.

3 2

22

2 2

1

x x dxx

3 2 3 22 2x x Ax Bx A C x B D 1A2B 0A C 2B D

Solve the system!!!

1, 2, 1, 0A B C D

Guided PracticeUse partial fractions to evaluate the integral.

3 2

22

2 2

1

x x dxx

22 2

21 1

x x dxx x

22 2 2

21 1 1

x x dxx x x

2 1

2

1 1ln 1 2 tan2 2 1

x x Cx

Guided PracticeSolve the given initial value problem.

22 1dy dx xy y 0 1y

2

1 21dy xdx

y y

2

1 21dy xdx

y y

22

111

A Bx Cy yy y

21 1A y Bx C y

21 A B y Cy A 0A B 0C 1A

1, 1, 0A B C

Guided PracticeSolve the given initial value problem.

22 1dy dx xy y 0 1y

2

1 21dy xdx

y y

2

1 21

y dy xdxy y

2 21ln ln 12

y y x C

Guided PracticeSolve the given initial value problem.

22 1dy dx xy y 0 1y

2 21ln ln 12

y y x C

Initial Condition: 2 21ln 1 ln 1 1 02

C

1 ln 22

C ln 2C

2 21ln ln 1 ln 22

y y x Solution: