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CALCULUS 1 PROBLEMS & SOLUTIONS
6.5.2
The Method
Of Partial Fractions
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Review
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 ax2 +
bx + c. A polynomial is said to be irreducible if it can't be
factored. For example, x2 – x – 2 can be
factored as (x + 1)(x – 2), while x2 + x + 1 can't be factored, so is irreducible.
Recall that ax2 + bx +
c is irreducible if b2 – 4ac < 0. Note that all linear functions are
irreducible.
A rational function is a ratio 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:
The first and last fractions must be identical, ie, equal to each other for all
x except x = –1
and x = 2
where the denominator is 0. Since their denominators are already identical, it
remains that their
numerators must be identical. Two polynomials are identical iff the
coefficients of like powers of x are
equal. It follows that we must have A + B = 5
(coefficients of x) and – 2A + B = –7
(coefficients of x0
= 1, which are the constant terms). Solving
the system of equations:
Another method of determining A and B is as follows. Multiplying both sides of Eq.
[1] by the
denominator x + 1 below A we obtain:
Case Of n Distinct Linear Factors
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.
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3. Partial Fractions – Quadratic Factors
Observe that the numerator of the first partial fraction is a constant, while
that of the second is a linear
function. Corresponding to a linear denominator we use a constant numerator,
and corresponding to a
quadratic denominator we use a linear numerator.
Now let's use the add-up-the-partial-fractions method to determine A, B, and C:
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 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:
which has no solutions. 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:
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.
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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 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.
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5. Partial Fractions – General Case
The following theorem of polynomial algebra summarizes the general case of the
partial-fraction
expansion of a rational function.
Theorem 1
¾¾¾¾¾¾¾¾¾¾¾¾
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:
¾¾¾¾¾¾¾¾¾¾¾¾
The proof of this theorem is omitted because it appropriately belongs to the
domain of polynomial
algebra. Here we simply use the theorem.
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6. The Method Of Partial Fractions
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. It's 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), 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).
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7. Using The Method Of Partial Fractions
Example 1
Find:
Partial fractions:
Note On Long Division
For example, given:
there's a “ trick ”,
which is simpler, to achieve the same thing:
Thus, if the degree of the numerator is equal to that of the denominator, we
can utilize some trick
equivalent to long division to achieve the same thing as long division. Another
example trick:
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Problems & Solutions
1. Find the following integrals.
Solution
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2. Calculate the following
integrals.
Solution
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3. Compute:
Solution
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4. Find:
Solution
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5. Find:
Solution
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