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Decompose $$\dfrac{2x}{1+x} $$

Looking at this case, it looks like any simple partial fraction. But it is trickly. This is how I attempted: $$\dfrac{2x}{1+x} = \dfrac{A}{1+x}$$

multiply by LCD $(1+x)$ to get $2x = A$

How to I reduce this to give me the value of $A$?

What if I do like this: $2x^1=Ax^0$ and conclude that $A = 2$?


by using long division: I am getting: $2+\dfrac{2}{x+1}$

$$================$$ @Ron Gordon et al: Ok I get what you mean. Finally, how then do I use the equation $1+\dfrac{1}{1+x}$ to come up with the partial fractions?

In my own thoughts I decided to eliminating the fraction in $1+\dfrac{1}{1+x}$

by multiply with LCD. I get:

$ \dfrac{1}{1}+\dfrac{1}{x+1}$ which gives $\dfrac{(x+1)+1}{(x+1)}$

$ \therefore$ our new equation to decompose is $$\dfrac{x+2}{(x+1)}$$

$\dfrac{(x+2)}{(x+1)} = $ .....is this equation now correct?

if so I proceed as below:

$\dfrac{x+2}{x+1} =\dfrac{A}{x+1}$

multiply both sides by LCD we get

$x+2=A$

to eliminate $x$ and it coefficient, let $x=0$ $$\therefore A = 2$$

$\therefore$ my solution is $ \dfrac{x+2}{x+1}=\dfrac{2}{x+1}$

Either am totally confused or the instructions here are not helping me understand this important concept?

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    $\begingroup$ Divide first, the degree of your numerator is the same as the degree of the denominator. $\endgroup$ – David Mitra Oct 1 '13 at 20:29
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    $\begingroup$ First you have to do a polynomial division with remainder to have the numerator of smaller degree than the denominator. $\endgroup$ – Daniel Fischer Oct 1 '13 at 20:29
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    $\begingroup$ Why would you want to eliminate $1\over {x+1}$? For all practical purposes you are done at that point. The PDF of $2x\over {x+1}$ is $2 - {2\over {x+1}}$. Where is your confusion? $\endgroup$ – JohanLiebert Oct 2 '13 at 11:22
  • $\begingroup$ ohhhhhhhhh! @JohanLiebert. That drove some sense into me. Thanks. $\endgroup$ – Sylvester Oct 2 '13 at 11:53
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$$\frac{x}{1+x} = \frac{1+x}{1+x} - \frac{1}{1+x} = 1-\frac{1}{1+x}$$

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  • $\begingroup$ I don't understand how you came up with $\dfrac{x}{1+x}$. Having derived the new equation after long division(above), how do I use it in formulating the partial equations? $\endgroup$ – Sylvester Oct 2 '13 at 7:53
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    $\begingroup$ @Sylvester: I am afraid I do not understand your question. I did not come up with that term - you gave it to me. I just neglected the factor of $2$ because I thought it was obvious that you just multiply my result by $2$ To get what you need. $\endgroup$ – Ron Gordon Oct 2 '13 at 7:56
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You should suppose the following as in general $$ \frac{x}{x+1}=f(x)+\frac{A}{1+x}, $$ where $f(x)$ is a polynomial (more precisely a constant in this case).

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