What technique turns $\frac{x}{(x-2)^2(x+1)}$ into $-\frac{1}{9x+9}+\frac{1}{9x-18}+\frac{2}{3(x-2)^2}$? I found this:

Let's rewrite the integrand so that it's easier to integrate:
  $$\dfrac{x}{(x-2)^2(x+1)} = -\dfrac{1}{9x+9}+\dfrac{1}{9x-18}+\dfrac{2}{3(x-2)^2}$$

I see how the expressions are equal, but I don't know how to get from the left to the right. There must be a technique that is used; could you tell me what the technique is called so I can read more about it or maybe you could explain it to me?
Thanks for your help.

The technique that is being used is called partial fraction decomposition. You'll get "tons" of hits if you Google the phrase.
As applied to this particular question, first note that
$$\dfrac{x}{(x-2)^2(x+1)} = -\dfrac{1}{9x+9}+\dfrac{1}{9x-18}+\dfrac{2}{3(x-2)^2}$$
$$= -\frac 19\cdot \frac 1{x + 1} + \frac 19 \cdot \frac{1}{x-2} + \frac 23\cdot \frac 1{(x-2)^2}\tag{1}$$
Now, back to the original fraction; let's go through how it is decomposed:
$$\dfrac{x}{(x-2)^2(x+1)} = \frac A{x+1} + \frac B{x-2} + \frac C{(x-2)^2}\tag{2}$$ Now the objective is to solve for $A, B, C$.
We now find that $$A(x-2)^2 + B(x+1)(x-2) + C(x+1) = x \tag{3}$$ $(3)$ is what we'd get if we multiplied each side of equation $(2)$ by the denominator of the right-hand side.
We can then solve for $A, B, C$ by expanding the left hand side of $(3)$, and equating coefficients from left and right, and obtaining a system of three equations in three unknowns. But in this case, another method works quite nicely.
Evaluate $(3)$ at $x = 2 \implies 3C = 2 \iff C= \frac 23$.
Evaluate $(3)$ at $x = -1 \implies 9A = -1\iff A =-\frac 19$.
Now, solve for $B$ using the found values of $A, C$, setting $x = 0$, say. Then evaluate. If you do this, you'll obtain $B = \frac 19$.
Now, replace $A, B, C$ in the right-hand side of equation $(2)$. Compare this with the right-most expression of equation $(1)$.
