Not a rearrangement. A decomposition, and it is what makes the thing integrable later.
The dark curve is 2x + 1(x − 1)(x + 2). Bring in the two pieces and watch them land on it exactly.
One fraction with two asymptotes, or two fractions with one each.
Write 2x + 1(x − 1)(x + 2) ≡ Ax − 1 + Bx + 2, multiply through by the denominator, and choose values of x that make one bracket vanish.
Choosing x to kill a bracket is the whole trick. Each substitution isolates one unknown, so you never solve simultaneous equations. The values to use are exactly the ones that make the original undefined, which is a pleasing bit of cheek and perfectly legitimate, because the identity holds for all other x.
Check at any third value. At x = 3, the original is 7/10 and the pieces give ½ + ⅕ = 7/10 ✓. One substitution confirms both constants.
The top must be of lower degree than the bottom. If it is not, divide first. At this level the denominator is two distinct linear factors and nothing worse.
Because 1x − 1 integrates to ln|x − 1| in one step, and the original integrates to nothing at all in the form it arrives in. Partial fractions is a technique from Topic 5 that happens to be taught in Topic 1.
1. For 5x − 1(x − 2)(x + 1), find A, the numerator over (x − 2).
2. For the same expression, find B.
3. Which x values are chosen to find A and B quickly?
Setting up the identity with the right form before substituting anything.
Choosing x to eliminate a bracket, rather than grinding out simultaneous equations.
Checking at a third value. It costs one line and it catches a sign error, which is the realistic failure here.
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