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Problem 75 Hard Difficulty

If $ a \neq 0 $ and $ n $ is a positive integer, find the partial fraction decomposition of
$$ f(x) = \frac{1}{x^n (x - a)} $$
[Hint: First find the coefficient of $ \frac{1}{(x - a)} $. Then subtract the resulting term and simplify what is left.]


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Related Courses

Calculus 2 / BC

Calculus: Early Transcendentals

Chapter 7

Techniques of Integration

Section 4

Integration of Rational Functions by Partial Fractions

Related Topics

Integration Techniques

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Integration Techniques - Intro

In mathematics, integration is one of the two main operations in calculus, with its inverse, differentiation, being the other. Given a function of a real variable, an antiderivative, integral, or integrand is the function's derivative, with respect to the variable of interest. The integrals of a function are the components of its antiderivative. The definite integral of a function from a to b is the area of the region in the xy-plane that lies between the graph of the function and the x-axis, above the x-axis, or below the x-axis. The indefinite integral of a function is an antiderivative of the function, and can be used to find the original function when given the derivative. The definite integral of a function is a single-valued function on a given interval. It can be computed by evaluating the definite integral of a function at every x in the domain of the function, then adding the results together.

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27:53

Basic Techniques

In mathematics, a technique is a method or formula for solving a problem. Techniques are often used in mathematics, physics, economics, and computer science.

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Watch More Solved Questions in Chapter 7

Problem 1
Problem 2
Problem 3
Problem 4
Problem 5
Problem 6
Problem 7
Problem 8
Problem 9
Problem 10
Problem 11
Problem 12
Problem 13
Problem 14
Problem 15
Problem 16
Problem 17
Problem 18
Problem 19
Problem 20
Problem 21
Problem 22
Problem 23
Problem 24
Problem 25
Problem 26
Problem 27
Problem 28
Problem 29
Problem 30
Problem 31
Problem 32
Problem 33
Problem 34
Problem 35
Problem 36
Problem 37
Problem 38
Problem 39
Problem 40
Problem 41
Problem 42
Problem 43
Problem 44
Problem 45
Problem 46
Problem 47
Problem 48
Problem 49
Problem 50
Problem 51
Problem 52
Problem 53
Problem 54
Problem 55
Problem 56
Problem 57
Problem 58
Problem 59
Problem 60
Problem 61
Problem 62
Problem 63
Problem 64
Problem 65
Problem 66
Problem 67
Problem 68
Problem 69
Problem 70
Problem 71
Problem 72
Problem 73
Problem 74
Problem 75

Video Transcript

Let's find the partial fraction in opposition. And here let's not that we're given is non zero. So here we can rewrite this. So for the first term X event, let's use what the author calls case, too. This is when you have repeated linear factors. So then a two over X squared and we go all the way up xn and then we'LL have a case One. This is a non repeatedly in your factor X minus a with good and multiply And both sides by the denominator on the left So on the right hand side and so on and we'LL have all the way up to a and X minus a and then we'LL have be accident Now the next step would be to go ahead and combine depending on the power of X So flu pullout xn one a one plus me Pull out a xn minus one. You get a two minus a one and so on. And when we plotted X day and minus a a and minus one and then finally the constant term and a So now we have comparing the constant terms We have a one on the left. We have a negative and and the right So that gives us a N equals negative one over, eh? And for instance, we can use that to find this. We know this is equal to zero because there's no X on the left. So we have a end here, Let me write it this way. And minus one equals a A and over, eh equals negative. A one over a square. Similarly, a N minus two equals a and minus one. Over, eh? So here, negative one. Over a cute. That's a negative up there and continuing in this fashion level. Me Go on to the next. Actually, let me take a step back here. That should have been in minus one. So what I had. Okay, so let me go. The next page, a one is a two over, eh? A three over a square. Just using the same formula repeatedly. Negative one over eight, then and then be equals negative. A one, because a one plus zero zero zero and then this is positive. One over. And so now that we've found all of our coefficients for the partial fraction to composition, just plug a moment. Negative one. An ex those are capital that was a one and then a two and minus one X squared and so on. Great. And then finally negative. One of rain and exit and and then R B was won over eight of the end, and that's her room was X minus a. There it is. That's our final answer.

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Related Topics

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Top Calculus 2 / BC Educators
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Numerade Educator

Heather Zimmers

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University of Michigan - Ann Arbor

Samuel Hannah

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Calculus 2 / BC Courses

Lectures

Video Thumbnail

01:53

Integration Techniques - Intro

In mathematics, integration is one of the two main operations in calculus, with its inverse, differentiation, being the other. Given a function of a real variable, an antiderivative, integral, or integrand is the function's derivative, with respect to the variable of interest. The integrals of a function are the components of its antiderivative. The definite integral of a function from a to b is the area of the region in the xy-plane that lies between the graph of the function and the x-axis, above the x-axis, or below the x-axis. The indefinite integral of a function is an antiderivative of the function, and can be used to find the original function when given the derivative. The definite integral of a function is a single-valued function on a given interval. It can be computed by evaluating the definite integral of a function at every x in the domain of the function, then adding the results together.

Video Thumbnail

27:53

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In mathematics, a technique is a method or formula for solving a problem. Techniques are often used in mathematics, physics, economics, and computer science.

Join Course
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