Partial Fractions: Solving Simple Linear Equations

AI Thread Summary
To solve the integration problem using partial fractions, the expression x^2 - 8 divided by x + 3 can be simplified through polynomial long division. The result is x - 3 plus a remainder of 1 over x + 3, leading to the equation x^2 - 8/(x + 3) = x - 3 + 1/(x + 3). Similarly, for x^2 + 1 divided by x - 1, it can be expressed as x + 1 plus a remainder of 2 over x - 1. Understanding how to manipulate the numerator and recognize patterns, such as the difference of squares, is crucial for these types of problems. Mastery of these techniques simplifies the integration process significantly.
Help1212
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Homework Statement


Hello I'm trying to figure out how to integrate with the use of partial fractions but I'm a bit stuck on something that should probably be simple.. but I can't see it clearly

How do you split up x2-8/x+3 and make it equal to (x-3) + 1/x+3

or for instance x2 +1/x-1 split up into x+1 +2/x-1

Homework Statement


Homework Equations


The Attempt at a Solution


They seem simple.. but i think I'm missing something here because I only see one linear denominator so I don't know how to apply the sequence.. or I'm overlooking something.. a bitof hellp please?
 
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It is basically "long division". To divide x^2- 8 by x+ 3, note that x divides into x^2 x times. Taking x times x+3, we get x^2+ 3x and subtract to get x^2- 8 - (x^2+ 3x)= -3x- 8. That is, x+ 3 divides into x^2- 8 x times with remainder -3x- 8. x divides into -3x -3 times. Taking -3 times x+ 3 we get -3x- 9 and subtracting that from -3x- 8 gives a remainder of +1. That tells you that
\frac{x^2- 8}{x+ 3}= x- 3+ \frac{1}{x+ 3}.

Another way to see that, simpler but not "rote", is to note that
\frac{x^2- 8}{x+ 3}= \frac{x^2- 9+ 1}{x+ 3}= \frac{(x- 3)(x+ 3)+ 1}{x+ 3}
= x- 3+ \frac{1}{x+ 3}
again.
 
Help1212 said:

Homework Statement


Hello I'm trying to figure out how to integrate with the use of partial fractions but I'm a bit stuck on something that should probably be simple.. but I can't see it clearly

How do you split up x2-8/x+3 and make it equal to (x-3) + 1/x+3

or for instance x2 +1/x-1 split up into x+1 +2/x-1

Knowing what to look for it is easy enough even when you have left out the brackets.

One useful thing for you would be to work backwards and check it that way - may give some insights.

One thing to look for is express your numerator as a [difference of two squares (check the formula for that) that has your denominator as factor] + something.
 
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