<s>[INST] How do I decompose this partial fraction case? 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?
 [/INST] $$\frac{x}{1+x} = \frac{1+x}{1+x} - \frac{1}{1+x} = 1-\frac{1}{1+x}$$
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