Do the algebra first AB & BC

Before touching a rational integrand, compare the degree of the numerator with the degree of the denominator.

1. Numerator degree $<$ denominator degree. The fraction is proper. Go on to the next section.
2. Numerator degree $\ge$ denominator degree. The fraction is improper, and it has to be divided before anything else is possible.

Long division turns an improper fraction into a polynomial plus a proper remainder. For $\int \frac{x^{2}}{x+1}\,dx$, dividing gives

$$\frac{x^{2}}{x+1} = x - 1 + \frac{1}{x+1}, \qquad \int \frac{x^{2}}{x+1}\,dx = \frac{x^{2}}{2} - x + \ln|x+1| + C.$$

Every piece on the right is something Topic 6.8 already handles, and nothing on the left was. That is the pattern for the whole topic.

Sometimes the division is a factoring in disguise, and spotting it saves the work: $\frac{x^{3}+x}{x^{2}+1} = \frac{x\left(x^{2}+1\right)}{x^{2}+1} = x$, so the integral is just $\frac{x^{2}}{2} + C$. Always look for a common factor before setting up the division.

What cannot happen is integrating the improper fraction directly. There is no rule turning $\frac{x^{2}}{x+1}$ into $\frac{x^{3}/3}{x^{2}/2 + x}$, and Topic 6.6's counterexample is the reason: integration does not distribute over division.

Once the fraction is proper, look at the denominator. If it is an irreducible quadratic, one with no real roots, completing the square converts it into the shape of an arctangent:

$$\int \frac{dx}{x^{2}+a^{2}} = \frac{1}{a}\arctan\!\left(\frac{x}{a}\right) + C.$$

For $\int \frac{dx}{x^{2}+4x+13}$, take half the coefficient of $x$, which is $2$, and square it:

$$x^{2}+4x+13 = \left(x^{2}+4x+4\right) + 9 = (x+2)^{2}+9,$$

so the integral is $\frac{1}{3}\arctan\!\left(\frac{x+2}{3}\right) + C$. Two details, and both cost marks.

1. The $\frac{1}{a}$ out front. Here $a = 3$, so a $\frac{1}{3}$ appears. It is the same reciprocal coefficient as in Topic 6.8, for the same chain-rule reason.
2. The shift goes inside. The argument is $\frac{x+2}{3}$, not $\frac{x}{3}$. Completing the square recentres the function at $x = -2$, and the arctangent has to be recentred with it.

Whether to complete the square is decided by the discriminant. If $b^{2} - 4ac < 0$ the quadratic never factors and the arctangent form is the target. If it does factor, the denominator comes apart into linear pieces, which on the AB path usually means a substitution and on the BC path means partial fractions in Topic 6.12.

A relative worth recognising: if the numerator is the derivative of the denominator, up to a constant, no completing is needed at all. $\int \frac{2x+4}{x^{2}+4x+13}dx$ is a substitution with $u$ the denominator, giving $\ln\left|x^{2}+4x+13\right| + C$.

Numerators split over a single denominator. Denominators never split at all.

$$\frac{a + b}{c} = \frac{a}{c} + \frac{b}{c} \quad\text{always}, \qquad \frac{a}{b + c} \neq \frac{a}{b} + \frac{a}{c} \quad\text{ever}.$$

The first is how $\int \frac{x^{2}+1}{x}dx$ becomes $\int\left(x + \frac{1}{x}\right)dx = \frac{x^{2}}{2} + \ln|x| + C$, which is a division you can do by inspection. The second is the error that turns $\frac{1}{x^{2}+1}$ into $\frac{1}{x^{2}} + 1$, and it changes the function before any calculus happens.

Three checks that keep this straight.

1. Test with numbers. $\frac{1}{2+3} = 0.2$ and $\frac{1}{2} + \frac{1}{3} \approx 0.833$. One line settles it permanently.
2. Split the numerator when the denominator is a single term, such as $x$ or $x^{2}$, and the whole thing usually collapses to power-rule pieces.
3. Never split when the sum is downstairs. A sum in the denominator is either factored, completed, or substituted for, and it is never distributed.

Almost every rational integrand on the AB exam falls into one of four cases, and the whole skill is telling them apart in one look.

1. Numerator is the derivative of the denominator, up to a constant. Substitute; the answer is a logarithm.
2. Improper fraction. Divide, then integrate the polynomial and whatever proper piece is left.
3. Proper, denominator an irreducible quadratic. Complete the square; the answer is an arctangent.
4. Denominator a single power of $x$. Split the numerator over it and use the power rule.

Two closing warnings.

An arctangent needs a constant numerator after preparation. $\int \frac{x}{x^{2}+9}dx$ is not an arctangent problem, it is a substitution giving $\frac{1}{2}\ln\left(x^{2}+9\right) + C$. If the numerator carries an $x$, check case 1 before reaching for case 3.

Preparation does not commute with integration. The rewriting happens to the integrand, before the integral sign is dealt with, and doing it afterwards is not a slower route to the same place. It is a different and wrong calculation.