How do I decompose this partial fraction case?

Question

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 $A = 2$?

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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?

Answer 1

$$\frac{x}{1+x} = \frac{1+x}{1+x} - \frac{1}{1+x} = 1-\frac{1}{1+x}$$

Answer 2

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).