Take the fraction apart first BC only

Two conditions have to hold before any decomposition is written down.

1. The fraction is proper: the numerator's degree is strictly less than the denominator's. If not, divide first, exactly as in Topic 6.10, and decompose only the remainder.
2. The denominator factors into distinct linear factors. That is the scope of this course: no repeated factors and no irreducible quadratics. An irreducible quadratic is completed instead, back in Topic 6.10.

Then each distinct linear factor gets one unknown constant over it:

$$\frac{1}{(x-3)(x+2)} = \frac{A}{x-3} + \frac{B}{x+2}.$$

The constants are the entire content of the method. Writing $\frac{1}{x-3} + \frac{1}{x+2}$ is not a decomposition with the constants left implicit; it is a different function. Adding those two gives $\frac{2x-1}{(x-3)(x+2)}$, whose numerator is nothing like $1$.

The numerator over a linear factor is a constant. Writing $\frac{Ax + B}{x - 3}$ is improper on its own, since the numerator's degree matches the denominator's, and it makes the system unsolvable.

Multiply through by the whole denominator to clear the fractions:

$$1 = A(x+2) + B(x-3).$$

This holds for every $x$, which gives two ways to finish.

1. Substitute the roots. Each root kills one term. Putting $x = 3$ gives $1 = 5A$, so $A = \frac{1}{5}$; putting $x = -2$ gives $1 = -5B$, so $B = -\frac{1}{5}$. This is fastest and it is what to use under time pressure.
2. Equate coefficients. Expand to $1 = (A+B)x + (2A - 3B)$ and match: $A + B = 0$ and $2A - 3B = 1$. Same answer, more algebra, and the only route when a repeated factor appears.

Notice that $B$ came out negative. Constants carry signs, and the expectation that a decomposition is a sum of positive pieces is a reliable source of lost marks. So is stopping once one constant is found.

Check by recombining. Putting $\frac{1/5}{x-3} - \frac{1/5}{x+2}$ over a common denominator gives $\frac{\frac{1}{5}(x+2) - \frac{1}{5}(x-3)}{(x-3)(x+2)} = \frac{1}{(x-3)(x+2)}$. That takes fifteen seconds and catches every sign error.

Each piece is a constant over a linear expression, and each one integrates to a logarithm:

$$\int \frac{k}{x - r}\,dx = k\ln|x - r| + C.$$

So the worked example finishes as

$$\int \frac{dx}{x^{2}-x-6} = \frac{1}{5}\ln|x-3| - \frac{1}{5}\ln|x+2| + C.$$

Three details.

1. The absolute value bars stay. The pieces are defined on both sides of each root, and $\ln$ of a negative number is not.
2. If the linear factor carries a coefficient, the reciprocal rule from Topic 6.8 applies: $\int \frac{dx}{2x+1} = \frac{1}{2}\ln|2x+1| + C$.
3. A single constant at the end. One $+ C$ covers the whole answer, not one per logarithm.

Logarithm laws can compress the result to $\frac{1}{5}\ln\left|\frac{x-3}{x+2}\right| + C$, which is equivalent and is not required. What is not permitted is the reverse move: $\ln|x-3| - \ln|x+2|$ is not $\ln|x - 3 - x - 2|$, and no law turns a difference of logarithms into a logarithm of a difference.

Once decomposed, a definite integral is Topic 6.7's routine. For $\int_{3}^{4} \frac{dx}{x^{2}-x-2}$, factor to $(x-2)(x+1)$, decompose to $\frac{1/3}{x-2} - \frac{1/3}{x+1}$, and evaluate:

$$\frac{1}{3}\Big[\ln|x-2| - \ln|x+1|\Big]_{3}^{4} = \frac{1}{3}\left[\left(\ln 2 - \ln 5\right) - \left(\ln 1 - \ln 4\right)\right] \approx 0.157.$$

The check that has to happen before any of that: are the roots of the denominator inside the interval? $\int_{0}^{3} \frac{dx}{x^{2}-x-2}$ cannot be evaluated this way at all, because the integrand blows up at $x = 2$. That is an improper integral, and Topic 6.13 takes it up.

Two closing notes on scope.

Partial fractions are the standard route into logistic differential equations in Unit 7, where $\frac{1}{P(M-P)}$ has to come apart before the variables can be separated. The technique is worth being fluent in for that reason as much as for this one.

If the denominator does not factor, this is the wrong tool. Check the discriminant first: a negative one means completing the square and an arctangent, which is Topic 6.10 rather than this one.