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Integration with partial fractions - help!

  • Thread starter Bensky
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Integration with partial fractions -- help!

Homework Statement


Here is the problem: http://img130.imageshack.us/img130/1673/integralthing.png [Broken]
The answer should be 4.


Homework Equations



N/A

The Attempt at a Solution



Here are my steps so far:

(5x^2 - 17x + 10) / ((x-1)^3 * (x+1)) = A/(x-1) + B/(x-1)^2 + C/(x-1)^3 + D/(x+1)
5x^2 - 17x + 10 = A(x-1)^2(x+1) + B(x-1)(x+1) + C(x+1) + D(x-1)^3

Substituting 1 for x gives:
5-17+10 = 2C
C = -1

Substituting -1 for x gives:
5(-1)^2 - 17(-1) + 10 = -8D
D = -4

Variables C and D are correct according to the answer given to me on the practice test.

Now I'm at the point where I want to find the other 2 variables, A and B. I'm not sure how to go about doing this. I talked to my professor and she said something about plugging in random values for x and somehow getting A and B that way, but it didn't make much sense to me after she explained it. I have to be able to do this for a test where I will not be allowed anything but the Apple OSX calculator (it's basically the same as the Windows Calculator), so the easiest method to do this would be appreciated.

Thank you,
Bensky

Homework Statement





Homework Equations





The Attempt at a Solution

 
Last edited by a moderator:

Answers and Replies

  • #2
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5,284


You're on the right track with the way you split up the separate fractions. The important thing is that your equation -- (5x^2 - 17x + 10) / ((x-1)^3 * (x+1)) = A/(x-1) + B/(x-1)^2 + C/(x-1)^3 + D/(x+1) -- has to hold for all values of x. That's what your teacher was talking about when he/she said to plug in random numbers. Presumably you have already tried x = 1 and x = -1. Another one to try is x = 0. Your choice for another one.
 
  • #3
82
0


You're on the right track with the way you split up the separate fractions. The important thing is that your equation -- (5x^2 - 17x + 10) / ((x-1)^3 * (x+1)) = A/(x-1) + B/(x-1)^2 + C/(x-1)^3 + D/(x+1) -- has to hold for all values of x. That's what your teacher was talking about when he/she said to plug in random numbers. Presumably you have already tried x = 1 and x = -1. Another one to try is x = 0. Your choice for another one.
Thanks Mark -- I actually figured this out on my own a minute or so before I saw this post.

For anyone else trying to do these types of problems: I ended up substituting x=2 and x=-2 to get the system of eq's (solved by elimination):
-3A + B = -15
3A + 3B = 3
-------------------
The A's already cancel, so...
4B = -12
B=-3 => A = 4

Maybe I should have screwed around a bit more before I posted here. :blushing:
 

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