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Get Full Access to Calculus: Early Transcendentals - 1 Edition - Chapter 7.4 - Problem 1e
Get Full Access to Calculus: Early Transcendentals - 1 Edition - Chapter 7.4 - Problem 1e

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# What kinds of functions can be integrated using partial

ISBN: 9780321570567 2

## Solution for problem 1E Chapter 7.4

Calculus: Early Transcendentals | 1st Edition

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Problem 1E

Problem 1E

What kinds of functions can be integrated using partial fraction decomposition?

Step-by-Step Solution:
Step 1 of 3

Problem 1E

Solution:-

Step1

we will assume that we have a rational function in which degree of p (x) < degree of q (x). If this is not the case, we can always perform long division.

For example, if we were given the fraction  .

We would perform long division to obtain

=x+

We then would apply partial fraction decomposition to .

Step2

(1)   q(x) is a product of distinct linear factors.

Let us assume that q (x) is a product of n distinct linear factors that is

q (x) =() ):::()

Then,

++-------------+

Finding the decomposition amounts to finding the coefficient  A1, ..., An. This can be done two different ways.

Step3

(2)q(x) is a product of linear factors, some being repeated

The factors which are not repeated will be decomposed as above.

Suppose that q (x) also contains  that is ax + b is repeated m times.

The decomposition for this factor will be

++------+

Step4

(c )q(x) is a product of distinct irreducible quadratic factors.

Recall that a term is called irreducible if it cannot be factored any further. Thus  is irreducible, so is. Be careful, is not considered a quadratic term. We must think of it as a linear term appearing twice. We can generalize what we did in the previous two cases as follows. Instead of thinking of linear factors, think that when we write the decomposition, the degree of the term in the numerator is 1 less than the degree of the term in the denominator. When we had linear factors in the denominator, it meant that we had to have terms of degree 0 in the numerator, that is we had constant terms. If the denominator contains irreducible quadratic factors, then the numerator will contain linear terms.

Step5

(d)q(x) is a product of irreducible quadratic factors, some being repeated

This is similar to step 3 , with linear terms in the numerator and quadratic terms in the denominator.

Step6

General Case: q(x) is a mixture of the above

Step7

Example

The idea behind this decomposition is that once the fraction is decomposed, we can integrate it.

Using partial decomposition

=

Now,

=

=    ( By using substitution.)

Step 2 of 3

Step 3 of 3

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