# Partial Fractions

## 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

## 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?

HallsofIvy
Homework Helper
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.

epenguin
Homework Helper
Gold Member

## 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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