Partial Fractions of 1/((3x)(5-x))

[312213]

Homework Statement

Find the partial fractions of 1/((3x)(5-x))

Homework Equations

The Attempt at a Solution

1/((3x)(5-x))=A/(3x)+B/(5-x)=(A(5-x)+B(3x))/((3x)(5-x))
1=A(5-x)+B(3x)
If x=0
1=A(5-0)+B(3*0)=5A => A=1/5
If x=5
1=A(5-5)+B(3*5)=B15 => B=1/15
So
1/((3x)(5-x))=1/(15x)+1/(15*(5-x))

This is what I think it is supposed to be but the answer is slightly different where the third term is negative, 1/((3x)(5-x))=1/(15x)-1/(15*(5-x)). I don't see where the negative sign comes from. It has been a while since I've done partial fractions so I'd like to know if this method is correct.

 

[flyingpig]

I suggest you take out that 3 in the bottom first.

That is

\frac{1}{3}\int \frac{dx}{x(5-x)}

 

[312213]

Pulling the 3 out only seems to make B=1/5 and everything else is the same. Can't see how that makes B negative either

 

[flyingpig]

Pulling the 3 out only seems to make B=1/5 and everything else is the same. Can't see how that makes B negative either

NO i just suggested it to make multiplication simpler. If you are in doubt, multiply and combine the fractions to see where you went wrong

 

[312213]

Just noticed that the negative came from switching the signs of (5-x) into (x-5) and that I didn't notice the answer had that the other way around too. Problem solved

 

[stevenb]

I don't see where the negative sign comes from. It has been a while since I've done partial fractions so I'd like to know if this method is correct.

Looks to me that you did it correctly. Is there any chance that the answer you were given has x-5 in the denominator rather than 5-x? If not, then the answer given was wrong, it seems.

So, the other day in another forum, I learned a new method of solving partial fractions by inspection. I've been doing them the traditional way, and someone pointed out the method of residues, which for some reason, I hadn't seen before. But, it hindsight is seems so obvious.

Anyway, if you don't have degenerate poles, the method is very quick. If you have degenerate poles the method is a little more involved, but not to bad.

Identify the poles ... in this case x=0 and x=5, the take the limit as x approaches each pole.

The limit of 1/((3x)(5-x))=A/(3x)+B/(5-x)as x goes to zero is

1/(5-x)=A=1/5

And, the limit of 1/((3x)(5-x))=A/(3x)+B/(5-x) as x goes to five is

1/(3x)=B=1/15

So, there is a quick way to check, or an even quicker way to cheat

 

[SammyS]

Notice that 1/(5-x) = -1/(x-5) so that \displaystyle \frac{1}{15x}+\frac{1}{15(5-x)}=\frac{1}{15x}-\frac{1}{15(x-5)}