# Antidifferentiation for a larger problem

1. Feb 21, 2007

### nealh149

1. The problem statement, all variables and given/known data
I'm trying to antidifferentiate a function in order to find the volume of a solid. The function comes out to be (2x+1)/(x^2+8x+16)

2. Relevant equations

N/A

3. The attempt at a solution
I've tried a multitude of things. I figured that partial fractions would work, but it turns out that since the denominator is a square that doesn't work. I'm not asking you to solve, just a nudge in the right direction

2. Feb 21, 2007

### cristo

Staff Emeritus
What do you mean by this? Try the partial fraction decomposition $$\frac{A}{(x+4)}+\frac{B}{(x+4)^2}$$

3. Feb 21, 2007

### nealh149

ALright. I don't understand though. Where did the third (x+4) come from. The denominator only factors to (x+4)(x+4)

4. Feb 21, 2007

### nealh149

Okay even when i try that I get

A(x+4)^2+B(x+4) = 2x+1

When I plug in to eliminate one of the variables (x=-4) then I get 0+0 = -7. I'm probably doing something wrong.

5. Feb 21, 2007

### cristo

Staff Emeritus
We know that the number of unknown constants must be equal to the degree of the denominator, and so the easiest way to set it up is to take the two fractions as A/(x+4) and B/(x+4)2

This is not correct: You should have the following expression: $$\frac{2x+1}{(x+4)^2}\equiv \frac{A}{(x+4)}+\frac{B}{(x+4)^2}$$ Multiplying this expression by (x+4)2 does not give what you have above.

6. Feb 21, 2007

### nealh149

Ok. I'll try that. Why did you use those denominators though, for the partials. Intuitively I would use x+4 and x+4 because they multiply together to get x^2 +8x+16

Thanks though

7. Feb 21, 2007

### nealh149

Alright. So I solve that equation and get 2x+1 = A(x+4) + B

A = 1 and B = -7

WHen I antidifferentiate I get ln(x+4) + 7/(x+4) Seems that I'm missing a factor of 2 in front of the first term. Sorry for the problems.

8. Feb 21, 2007

### nealh149

nevermind on that last part, i found a mistake. but the previous question is still troubling me

9. Feb 21, 2007

### cristo

Staff Emeritus
Well, if we use (x+4) as the only denominator we obtain something like $$\frac{2x+1}{(x+4)^2}\equiv \frac{A}{x+4}+\frac{B}{x+4}=\frac{A+B}{x+4}$$; notice that this can never be in the required form (since the denominator will always be linear). Thus, we take the two fractions as (x+4) and (x+4)2, so that the fraction looks like$$\frac{A}{x+4}+\frac{B}{(x+4)^2}=\frac{A(x+4)+B}{(x+4)^2}$$ which is of the required form.

10. Feb 21, 2007

### nealh149

Alright that makes sense. I'm trying to solve that equation now. I can solve for B=-7, but I can't solve for A because I have two variables: x and A.

11. Feb 21, 2007

### cristo

Staff Emeritus
Just put, say, x=0.

12. Feb 21, 2007

### nealh149

Oh nvm. I got it. I have the problem solved now. I really appreciate the help cristo, you've enhanced my understanding of partial fractions.

13. Feb 21, 2007

### cristo

Staff Emeritus
You're very welcome; glad to be of help!