Calculus Of One Real Variable – By Pheng Kim Ving

10.7 
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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
(its graph is a
line). A quadratic function is a polynomial of degree 2, ie a function of
the form ax^{2} + bx + c. A realvalued polynomial is
said to be irreducible if it can't be factored. Note that all realvalued
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.
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2. Partial
Fractions – Linear Factors

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:
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:
The constants A_{i}'s, i = 1, 2, ..., n, can be determined by the
addupthepartialfractions method or the limitprocedure
method as in the above example, where n
= 2.
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3. Partial Fractions – Quadratic Factors 
Consider as another example the rational function:
Solving the system of equations:
Here, the limitprocedure 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/(x^{2} + 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/(x^{2} + 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:
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
A_{i}'s, i
= 1, 2, ..., m, B_{j}'s, and C_{j}'s,
j = 1, 2, ..., n, can be determined by the
addupthepartialfractions method as in the
above example, where m
= n = 1.
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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
x^{2} + 5, with denominators'
exponents increasing from 1 to 3.
The constants A_{1}, A_{2}, A_{3}, A_{4}, B, C_{1}, C_{2}, C_{3}, D_{1}, D_{2}, and D_{3} can be determined by the
addupthepartialfractions
method.
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5. Partial Fractions – General Case 
The following theorem of polynomial algebra summarizes the
general case of the partialfraction expansion of a rational
function.
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.
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6. The Method Of Partial Fractions 
Example 6.1
Find:
EOS
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:
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).
ii. Factor the denominator Q(x) into linear and/or irreducible quadratic factors.
iii. Perform the partialfraction 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).
For example, given:
Problems & Solutions 
1. Calculate the following integrals.
Solution
2. Compute the following integrals.
Solution
3. Evaluate:
Solution
where C = (1/2)C_{1}.
4. Find:
Solution
5. Calculate:
Solution
Let u = e^{x}. Then du = e^{x} dx = u dx, yielding dx = (du)/u. So:
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