Partial Fractions turn out wrong

[authgeek]

Here's (what I think is) a step in partial fractions that I don't understand:
http://apthtml.com/images/partialfrac.png

I'm taking the regular partial fractions steps and I keep ending up with 1 / (1 - x) rather than 1 / (x - 1). What am I doing wrong?

 

[slider142]

Here's (what I think is) a step in partial fractions that I don't understand:
http://apthtml.com/images/partialfrac.png

I'm taking the regular partial fractions steps and I keep ending up with 1 / (1 - x) rather than 1 / (x - 1). What am I doing wrong?

Are the signs the same as well? Remember, -1/(x-1) = 1/(1-x).

 

[sylas]

Here's (what I think is) a step in partial fractions that I don't understand:
http://apthtml.com/images/partialfrac.png

I'm taking the regular partial fractions steps and I keep ending up with 1 / (1 - x) rather than 1 / (x - 1). What am I doing wrong?

You don't show your steps, so I don't know what you are doing wrong. But try this. Let x be 0.5.

$$\begin{align*}<br /> x &= 0.5 \\<br /> \frac{1}{x(1-x)} &= \frac{1}{0.5 \times (1 - 0.5)} \\<br /> & = \frac{1}{0.5 \times 0.5} \\<br /> &= \frac{1}{0.25} \\<br /> & = 4 \\<br /> \frac{1}{x} - \frac{1}{x - 1} &= \frac{1}{0.5} - \frac{1}{0.5 - 1} \\<br /> & = 2 - -2 \\<br /> & = 4 \\<br /> \frac{1}{x} - \frac{1}{1 - x} &= \frac{1}{0.5} - \frac{1}{1 - 0.5} \\<br /> & = 2 - 2 \\<br /> & =0<br /> \end{align*}$$​

So you can see the second form has to be wrong.

If you can't see which step of your argument is false, just substitute 0.5 for x at all points, and see at which step you get the wrong answer. Stare at that step until you are enlightened.

Cheers -- sylas

 

[authgeek]

slider: that's definitely the case here, subtle switching of signs and a reversal rule I wasn't aware of. I suppose knowing these rules just comes with doing a lot of math.

silas: I appreciate the detailed answer, and particularly your advice to "Stare at that step until you are enlightened." I feel like I do that quite often... and it works!