Partial fraction $\frac{x}{(1+x)^2}$ How do you compute $\dfrac{x}{(1+x)^2}$ using partial fractions? The reason I ask is because when I try to solve it I keep getting an impossible $A, B$.
$A(1+x) + B(1+x) = x$
$A + Ax + B + Bx = x$
$(A+B)x = 1$
$(A+B) = 0$
However a practice problem I'm working on implies this can be solved by partial fractions, but doesn't go into detail. 
 A: Raskolnikov's answer is a pretty slick way to do this, but if you wanted a more general algorithm for problems like this, here you go.
Whenever the denominator is some linear term squared, like it is here, since it's $(1 + x)^2$, then you set up the partial fraction this way:
$$\frac{x}{(1 + x)^2} = \frac{A}{1 + x} + \frac{B}{(1 + x)^2}$$
Then solving this is just a matter of multiplying both sides by $(1 + x)^2$.
A: $$\frac{x}{(1+x)^2}=\frac{1+x-1}{(1+x)^2}=\frac{1}{(1+x)}-\frac{1}{(1+x)^2} $$
A: $$\frac{x}{(1 + x)^2} = \frac{A}{1 + x} + \frac{B}{(1 + x)^2}$$
$$x=A(1+x)+B $$
put $x=-1$ gives $B=-1$
and compare coefficient of x from both side gives $A=1$
so answer is $$\frac{x}{(1 + x)^2} = \frac{1}{1 + x} - \frac{1}{(1 + x)^2}$$
A: In general, whenever you have an expression of the form
$$\dfrac{P(x)}{(x+a)^n}$$ where $P(x)$ is of degree less than $n$ and you want to write it into partial fraction, then you need to proceed as follows.
$$\dfrac{P(x)}{(x + a)^n} = \dfrac{A_1}{x+a} + \dfrac{A_2}{(x+a)^2} + \dfrac{A_3}{(x+a)^3} + \cdots + \dfrac{A_{n-1}}{(x+a)^{n-1}} + \dfrac{A_n}{(x+a)^n} \tag{$\star$}$$
and then find the coefficients $A_1, A_2,\ldots, A_n$. Note that from $(\star)$, we have
$$P(x) = A_1(x+a)^{n-1} + A_2 (x+a)^{n-2} + A_3(x+a)^{n-3} + \cdots + A_n$$
From this, we have $A_n = P(-a)$ and in general, $$A_k = \left.\dfrac{d^{n-k} P(x)}{dx} \right \vert_{x=-a}$$
In your case, though it is relatively simple since
$$\dfrac{x}{(1+x)^2} = \dfrac{1+x-1}{(1+x)^2} = \dfrac1{(1+x)} - \dfrac1{(1+x)^2}$$
