section 8.4a. a flashback to section 6.5… we evaluated the following integral: this expansion...
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Section 8.4aPARTIAL FRACTIONS
A flashback to Section 6.5…We evaluated the following integral:
100
100dP
P P1 1
100dP
P 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 A
x 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 C
x 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 3
2 3
xdx
x x
2
5 3
2 3
x
x x
5 3
1 3
x
x x
1 3
A B
x 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 3
2 3
xdx
x x
1 3
A Bdx
x x 2 3
1 3dx
x x 2ln 1 3ln 3x x C
Guided PracticeUse partial fractions to evaluate the integral.
26 7
2
xdx
x
26 7
2
x
x
22 2
A B
x 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 7
2
xdx
x
22 2
A Bdx
x x
26 5
2 2dx
x x
26 5 2
2
dxx dx
x
1
6ln 2 5 2x x C
Guided PracticeUse partial fractions to evaluate the integral.
3 2
2
2 4 3
2 3
x x xdx
x 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 32
2 3
xx dxx x
Guided PracticeUse partial fractions to evaluate the integral.
3 2
2
2 4 3
2 3
x x xdx
x x
2
5 32
2 3
xx dxx x
2
5 32
2 3
xxdx dx
x x
2 3
21 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 xdx
x
3 2
22
2 2
1
x x
x
22 21 1
Ax B Cx D
x 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 xdx
x
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 xdx
x
22 2
2
1 1
x xdx
x x
22 2 2
2
1 1 1
x xdx
x 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
12
1dy xdx
y y
2
12
1dy xdx
y y
22
1
11
A Bx C
y 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
12
1dy xdx
y y
2
12
1
ydy xdx
y y
2 21ln ln 1
2y y x C
Guided PracticeSolve the given initial value problem.
22 1dy dx xy y 0 1y
2 21ln ln 1
2y y x C
Initial Condition: 2 21ln 1 ln 1 1 0
2C
1ln 22
C ln 2C
2 21ln ln 1 ln 2
2y y x Solution: