Recall from algebra that a linear function is a polynomial of degree 1, ie a function of the form ax + b (its graph is a
line). A quadratic function is a polynomial of degree 2, ie a function of the form ax2 + bx + c. A real-valued polynomial is
said to be irreducible if it can't be factored. Note that all real-valued linear functions are irreducible.
line). A quadratic function is a polynomial of degree 2, ie a function of the form ax2 + bx + c. A real-valued polynomial is
said 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 to
a substitution method. If that's the case, it'll be expressed as a sum of simpler fractions, known as partial fractions, which
are easier to integrate.
a substitution method. If that's the case, it'll be expressed as a sum of simpler fractions, known as partial fractions, which
are easier to integrate.
Consider, for example, the rational function:
Indeed it's correct.
Another method of determining A and B is as follows. Multiplying both sides of Eq. [2.1] by the denominator x + 1 below
A we obtain:
A we obtain:
Case 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 a
product of n distinct linear factors, say:
product of n distinct linear factors, say:
The constants Ai's, i = 1, 2, ..., n, can be determined by the add-up-the-partial-fractions method or the limit-procedure
method as in the above example, where n = 2.
method as in the above example, where n = 2.
3. Partial Fractions – Quadratic Factors |
Consider as another example the rational function:
Solving the system of equations:
Here, the limit-procedure method can be used to determine A, but there's no simple way to use it to determine B or C.
Remark 3.1
You may ask why we don't use a constant numerator for a partial fraction with a quadratic denominator to make things
simpler, like this: A/(x + 2) + B/(x2 + x + 1). Well, let's see:
simpler, 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 can be expanded to
A/(x + 2) + B/(x2 + x + 1). That's the answer to the question.
A/(x + 2) + B/(x2 + x + 1). That's the answer to the question.
Case 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 a
product of m distinct linear factors and n distinct irreducible quadratic factors, say:
product of m distinct linear factors and n distinct irreducible quadratic factors, say:
Again corresponding to a linear denominator we use a constant numerator and corresponding to a quadratic denominator
we use a linear numerator. That is, the degree of the numerator is less than that of the denominator by 1. The constants
Ai's, i = 1, 2, ..., m, Bj's, and Cj's, j = 1, 2, ..., n, can be determined by the add-up-the-partial-fractions method as in the
above example, where m = n = 1.
we use a linear numerator. That is, the degree of the numerator is less than that of the denominator by 1. The constants
Ai's, i = 1, 2, ..., m, Bj's, and Cj's, j = 1, 2, ..., n, can be determined by the add-up-the-partial-fractions method as in the
above example, where m = n = 1.
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, with denominators having
exponents increasing from 1 to 4. There's only 1 partial fraction corresponding to x – 3, and there are 3 corresponding to
x2 + 5, with denominators' exponents increasing from 1 to 3.
exponents increasing from 1 to 4. There's only 1 partial fraction corresponding to x – 3, and there are 3 corresponding to
x2 + 5, with denominators' exponents increasing from 1 to 3.
The constants A1, A2, A3, A4, B, C1, C2, C3, D1, D2, and D3 can be determined by the add-up-the-partial-fractions
method.
method.
5. Partial Fractions – General Case |
The following theorem of polynomial algebra summarizes the general case of the partial-fraction expansion of a rational
function.
function.
Theorem 5.1
Let Q(x) be a polynomial. Then Q(x) can be factored into a product of a constant, linear factors, and irreducible quadratic factors, as follows:  |
The proof of this theorem is omitted because it appropriately belongs to the domain of polynomial algebra. Here we simply
utilize the theorem.
utilize the theorem.
6. The Method Of Partial Fractions |
Example 6.1
Find:
Solution
EOS
Procedure
Suppose we are to find the integral:
If we don't know how to do it, we decompose P(x)/Q(x) into a sum of partial fractions and integrate the resulting
expression. This technique is called the method of partial fractions. Its procedure is summarized as follows:
expression. This technique is called the method of partial fractions. Its procedure is summarized as follows:
i. If the degree of P(x) is greater than or equal to that of Q(x), use polynomial long division to divide 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 the quotient, R(x) is the
remainder, and the degree of R(x) is less than that of Q(x).
obtain P(x)/Q(x) = q(x) + R(x)/Q(x) (from P(x) = q(x)Q(x) + R(x)), where q(x) is the quotient, 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:
Problems & Solutions |
1. Calculate the following integrals.
Solution
2. Compute the following integrals.
Solution
3. Evaluate:
Solution
where C = (1/2)C1.
4. Find:
Solution
5. Calculate:
Solution
Let u = ex. Then du = ex dx = u dx, yielding dx = (du)/u. So:
No comments:
Post a Comment