# Help with integration via partial fractions

1. Apr 23, 2006

### Sisyphus

I'm supposed to integrate this using partial fractions:

$$\int\frac{1}{(x-1)^2(x+1)} \ dx$$

I've started to split the integrand into more readily integrated fractions by stating...

$$\frac{A}{(x-1)}+\frac{B}{(x-1)^2}+\frac{C}{(x+1)} = \frac{1}{(x-1)^2(x+1)}$$

combining the fractions via addition, and making the numerators equal to eachother (because we have common denominators on each side), I get

$$x^3(A+C)+X^2(-A+B-3C)+X(-A+3C)+A-B-C=1$$

From this equation, I can deduce the following:

1. $$A+C=0$$

2. $$-A+B-3C=0$$

3. $$-A+3C=0$$

4. $$A-B-C=1$$

The problem is, the results I get from these equations contradict eachother. ie, if add equations 1 and 3 together, I get that C=0, but if I add equations 2 and 3 together, I get that $$C=\frac{-1}{4}$$

What in god's name am I doing wrong? I've done the calculations a number of times, and am pretty sure that my algebra isn't off...any help?

I have a hunch that my problem comes from the way I split up the fraction into three parts..

Last edited: Apr 23, 2006
2. Apr 23, 2006

### nrqed

There is no way to get C=-1/4 using only eqs 2 and 3. if you *add* them, C drops out. If you subtract them, B remains there

3. Apr 23, 2006

### dav2008

Well the problem off the bat is that you have 4 equations and 3 unknowns.

The next (and directly related) problem is that I'm not quite sure where you're getting your x3 term from...

Looking at your second line: $$\frac{A}{(x-1)}+\frac{B}{(x-1)^2}+\frac{C}{(x+1)} = \frac{1}{(x-1)^2(x+1)}$$I'm not sure why you have B divided by (x-1)2.

Last edited: Apr 23, 2006
4. Apr 23, 2006

### nrqed

Your mistake is that you multiplied A by (x-1)^2 (x+1). You must only multiply it by (x-1)*(x+1)

EDIT: you made the same mistake with the others. You multiply by too many factors. For example, B must get multiplied by (x+1) only

Last edited: Apr 23, 2006
5. Apr 23, 2006

### nrqed

If I recall, in partial fractions if there is a term of higher power, one must include terms all the way to the highest power hence A/(x-1) +B/(x-1)^2

6. Apr 23, 2006

### Sisyphus

Well, my reasoning for splitting up the fractions the way I did, was because my teacher and my textbook both told me that when the denominator is comprised of linear factors, some of which are repeated, I should go about the problem the following manner:

$$\frac{6x+7}{(x+2)^2} = \frac{A}{x+2} + \frac{B}{(X+2)^2}$$

the above is an example straight out of the textbook.

The truth is, I was given very little justification for having to split up the fraction in this way, and therefore have a very muddled understanding of the whole process in general. My attempt at trying to solve the initial question basically represents me trying to piece all the rules I got for partial fractions in a disjointed manner.

How do you guys suggest I go about tacking the original problem posted in my first post? Trigonometric substitution is out of question right now, since I haven't covered it yet. Perhaps I should try and use integration by parts? I don't know. I'm honestly pretty stumped with this one. I think all I need is a nudge in the right direction here.

7. Apr 23, 2006

### dav2008

$$\int\frac{1}{(x-1)^2(x+1)} \ dx$$ is the same as
$$\int\frac{1}{(x-1)(x-1)(x+1)} \ dx$$

I guess you're right that it doesn't really matter what the denominator of your fractions are as long as you multiply A, B, and C by the proper terms.

In your example of $$\frac{6x+7}{(x+2)^2} = \frac{A}{x+2} + \frac{B}{(X+2)^2}$$ you would have your equation as $$A(x+2)+B=6x+7$$

8. Apr 23, 2006

### nrqed

No, keep going! You did write the correct decomposition. It is just when you put them over a common denominator that you multiplied by two many factors.

For example ${A \over (x-1)} = {A (x-1)(x+1) \over (x-1)^2 (x+1) }$ and so on. You see? The idea is to bring all your terms over the common factor (x-1)^2(x+1) so that now you can set all this equal to your initial expression. Your mistake was to multiply A, B and C all by too many factors.

Pat

9. Apr 23, 2006

### Sisyphus

!!!

Good lord, I feel like a moron :rofl:

thanks!!