AP Calculus BC

Partial Fractions vs Long Division

Check the degrees first: if the numerator degree is at least the denominator degree, long division runs first and partial fractions applies only to the proper remainder. Partial fractions on an improper fraction produces a contradiction, because constant numerators cannot rebuild a numerator of that degree.

Partial fractions

Use when: The numerator degree is strictly below the denominator degree and the denominator factors into distinct linear pieces.

Long division

Use when: The numerator degree matches or exceeds the denominator degree, so divide first and pass the leftover fraction along.

Side by side

	Partial fractions	Long division
Runs when	Numerator degree is below denominator degree	Numerator degree is greater than or equal to denominator degree
Order of operations	Second, applied to the proper remainder	First, before anything else is attempted
What it produces	A sum of fractions with constant numerators	A polynomial plus a proper fraction
Denominator requirement	Factors into distinct linear factors for AP purposes	None, any denominator divides
On $\frac{x^2}{x^2 - 1}$	Applies only afterwards, to $\frac{1}{x^2 - 1}$	Produces $1 + \frac{1}{x^2 - 1}$

The degree comparison sets the order of operations and takes two seconds. Numerator degree below denominator degree means the fraction is proper and partial fractions can start straight away. Numerator degree matching or exceeding the denominator degree means the fraction is improper, and long division has to peel off a polynomial first, leaving a proper remainder for partial fractions to work on.

\frac{x^2}{x^2 - 1} = 1 + \frac{1}{x^2 - 1} = 1 + \frac{1}{2}\cdot\frac{1}{x - 1} - \frac{1}{2}\cdot\frac{1}{x + 1}

That chain delivers the integral in one pass: $\int \frac{x^2}{x^2 - 1}\,dx = x + \frac{1}{2}\ln\lvert x - 1 \rvert - \frac{1}{2}\ln\lvert x + 1 \rvert + C$ . Differentiating the answer returns $1 + \frac{1}{x^2 - 1}$ , which is the original integrand, so the division step is no tidy-up. The lone $x$ at the front of the antiderivative comes from it and from nowhere else.

What going straight to partial fractions does

Try $\frac{x^2}{(x-1)(x+1)} = \frac{A}{x-1} + \frac{B}{x+1}$ and clear denominators to reach $x^2 = A(x+1) + B(x-1)$ . Setting $x = 1$ gives $A = \frac{1}{2}$ and $x = -1$ gives $B = -\frac{1}{2}$ , but $x = 0$ then demands $0 = A - B = 1$ . The contradiction is the signal that the division was skipped, since a degree two left side can never match a degree one right side.

Frequently asked questions

Do I need long division before partial fractions?

Only when the fraction is improper, and then always. Compare the two degrees before writing anything: if the top wins or ties, divide, and if the top loses, go straight to partial fractions.

How do I know if a rational function is improper?

Look at the highest power on top and the highest power on the bottom. $\frac{x^3 + 1}{x^2 - 4}$ is improper and $\frac{x + 1}{x^2 - 4}$ is proper, while $\frac{x^2}{x^2 - 1}$ is improper too, because equal degrees still count as improper.

What partial fractions do I need for AP Calculus BC?

Only denominators that factor into distinct linear factors, which is the scope the course description sets. Repeated linear factors and irreducible quadratic factors sit outside the exam, so a denominator such as $x^2 - 5x + 6$ is fair game and $(x-2)^2$ is not.

In the CED: Unit 6: Integration and Accumulation