the method of partial fractions
TRANSCRIPT
-
8/3/2019 The Method of Partial Fractions
1/19
The Method Of Partial Fractions
Return To Contents Go To Problems & Solutions
1. Notes
Recall from algebra that a linear function is a polynomial of degree 1, ie a function of the form ax + b (itsgraph is aline ). A quadratic function is a polynomial of degree 2, ie a function of the form ax 2 + bx + c. A real-valued polynomial issaid to be irreducible if it can't be factored. Note that all real-valued linear functions are irreducible.
A rational function is a ratio or fraction P ( x) / Q( x) where P ( x) and Q( x) are polynomials.
In this section we're concerned with the integration of rational functions. A rational function may not readily lend itself toa substitution method. If that's the case, it'll be expressed as a sum of simpler fractions, known as partialfractions, whichare easier to integrate.
Go To Problems & Solutions Return To Top Of Page
2 . Partial Fractions Linear Factors
Consider, for example, the rational function:
-
8/3/2019 The Method of Partial Fractions
2/19
Indeed it's correct.
Another method of determining A and B is as follows. Multiplying both sides of Eq. [2.1] by thedenominator x + 1 below A we obtain:
C ase Of n Distinct Linear Factors
In general, if the degree of the numerator P ( x) is less than that of the denominator Q( x) and if Q( x)factors into aproduct of n distinct linear factors, say:
-
8/3/2019 The Method of Partial Fractions
3/19
-
8/3/2019 The Method of Partial Fractions
4/19
Solving the system of equations:
-
8/3/2019 The Method of Partial Fractions
5/19
Here, the limit-procedure method can be used to determine A, but there's no simple way to use it todetermine B or C .
R emark 3 .1
You may ask why we don't use a constant numerator for a partial fraction with a quadratic denominatorto make thingssimpler, like this: A / ( x + 2) + B / ( x2 + x + 1). Well, let's see:
which has no solutions. That is, There are no constants A and B such that the given rational function canbe expanded to A / ( x + 2) + B / ( x2 + x + 1). That's the answer to the question.
C ase Of m Distinct Linear Factors And n Distinct Quadratic Factors
In general, if the degree of the numerator P ( x) is less than that of the denominator Q( x) and if Q( x)factors into aproduct of
mdistinct linear factors and
ndistinct irreducible quadratic factors, say:
Again corresponding to a linear denominator we use a constant numerator and corresponding to aquadratic denominatorwe use a linear numerator. That is, the degree of the numerator is less than that of the denominator by1. The constants Ai's, i = 1, 2, ..., m, B j's, and C j's, j = 1, 2, ..., n, can be determined by the add-up-the-partial-fractionsmethod as in theabove example, where m = n = 1.
Go To Problems & Solutions Return To Top Of Page
-
8/3/2019 The Method of Partial Fractions
6/19
4 . Partial Fractions Multiplicity
Consider, for example, the rational function:
Since the multiplicity of the factor x is 4, there are 4 partial fractions corresponding to x, withdenominators havingexponents increasing from 1 to 4. There's only 1 partial fraction corresponding to x 3, and there are 3corresponding to x2 + 5, with denominators' exponents increasing from 1 to 3.
The constants A1, A2, A3, A4, B, C 1, C 2, C 3, D1, D2, and D3 can be determined by the add-up-the-partial-fractionsmethod.
Go To Problems & Solutions Return To Top Of Page
5 . Partial Fractions General C ase
The following theorem of polynomial algebra summarizes the general case of the partial-fractionexpansion of a rationalfunction.
Theorem 5 .1
Let Q( x) be a polynomial. Then Q( x) can be factored into a product of a constant, linear factors, and irreduciblequadratic factors, as follows:
-
8/3/2019 The Method of Partial Fractions
7/19
The proof of this theorem is omitted because it appropriately belongs to the domain of polynomialalgebra. Here we simplyutilize the theorem.
Go To Problems & Solutions Return To Top Of Page
6 . The Method Of Partial Fractions
Example 6 .1
Find:
-
8/3/2019 The Method of Partial Fractions
8/19
-
8/3/2019 The Method of Partial Fractions
9/19
i. If the degree of P ( x) is greater than or equal to that of Q( x), use polynomial long division todivide P ( x) by Q( x) to
obtain P ( x) / Q( x) = q( x) + R( x) / Q( x) (from P ( x) = q( x)Q( x) + R( x)), where q( x) is thequotient, R( x) is the
remainder, and the degree of R( x) is less than that of Q( x).
ii. Factor the denominator Q( x) into linear and/or irreducible quadratic factors.
iii. Perform the partial-fraction expansion on P ( x) / Q( x), or on R( x) / Q( x) if part i is carried out.
iv. Integrate the resulting expression of P ( x) / Q( x).
Note On Long Division
For example, given:
Return To Top Of Page
Problems & S olutions
-
8/3/2019 The Method of Partial Fractions
10/19
1. Calculate the following integrals.
S olution
-
8/3/2019 The Method of Partial Fractions
11/19
-
8/3/2019 The Method of Partial Fractions
12/19
Return To Top Of Page
2 . Compute the following integrals.
S olution
-
8/3/2019 The Method of Partial Fractions
13/19
-
8/3/2019 The Method of Partial Fractions
14/19
-
8/3/2019 The Method of Partial Fractions
15/19
Return To Top Of Page
3 . Evaluate:
S olution
-
8/3/2019 The Method of Partial Fractions
16/19
-
8/3/2019 The Method of Partial Fractions
17/19
where C = (1/2) C 1.
Return To Top Of Page
4 . Find:
S olution
-
8/3/2019 The Method of Partial Fractions
18/19
-
8/3/2019 The Method of Partial Fractions
19/19
Let u = e x. Then d u = e x dx = u dx, yielding dx = ( d u )/ u . So: