Topic 6.12 Notes – Integrating Using Linear Partial Fractions

What linear partial fractions are

A rational function looks like
$$\frac{P(x)}{Q(x)}$$ where both $P(x)$ and $Q(x)$ are polynomials.

This topic applies when:

• The degree of $P(x)$ is less than the degree of $Q(x)$.
• The denominator factors into distinct linear factors, like $(x-2)(x+5)$.
• No repeated factors and no irreducible quadratics (those are different setups).

Example structure:

$$\frac{P(x)}{(x-a)(x-b)} = \frac{A}{x-a} + \frac{B}{x-b}.$$

Why this works: each piece integrates easily because
$$\int \frac{1}{x-a}\,dx = \ln|x-a| + C.$$

So the big idea is:

Complicated rational function → sum of simple log integrals

When this method applies

You should immediately think partial fractions when:

• You see a rational function.
• The denominator factors into nonrepeating linear factors.
• The numerator’s degree is smaller.

If the fraction is improper (top degree ≥ bottom degree), do polynomial long division first, then decompose what remains.

If the denominator doesn’t factor over the reals, or has repeated factors, that’s still partial fractions but not this specific linear distinct case.

On non-calculator parts of tests, factoring correctly is often the real challenge. Slow down there.

The decomposition setup

Suppose you need to integrate

$$\frac{3x+4}{(x-1)(x+2)}.$$

Because the denominator has two distinct linear factors, we write:

$$\frac{3x+4}{(x-1)(x+2)} = \frac{A}{x-1} + \frac{B}{x+2}.$$

If there were three factors, you’d use three fractions:

$$\frac{P(x)}{(x-a)(x-b)(x-c)} = \frac{A}{x-a} + \frac{B}{x-b} + \frac{C}{x-c}.$$

Every linear factor gets its own constant numerator.

Solving for the constants

Using the example:

$$\frac{3x+4}{(x-1)(x+2)} = \frac{A}{x-1} + \frac{B}{x+2}.$$

Step 1: Clear denominators

Multiply both sides by $(x-1)(x+2)$:

$$3x+4 = A(x+2) + B(x-1).$$

Now it’s a polynomial identity.

Step 2: Solve using substitution

Plug in values that cancel terms.

• Let $x=1$:

$$3(1)+4 = A(3) + B(0)$$

$$7 = 3A \Rightarrow A = \frac{7}{3}.$$

• Let $x=-2$:

$$3(-2)+4 = A(0) + B(-3)$$

$$-6+4 = -3B$$

$$-2 = -3B \Rightarrow B = \frac{2}{3}.$$

Now rewrite:

$$\frac{3x+4}{(x-1)(x+2)} = \frac{7/3}{x-1} + \frac{2/3}{x+2}.$$

Substitution is usually fastest on exams.

Integrating the decomposed form

Now integrate term by term:

$$\int \frac{7/3}{x-1} dx + \int \frac{2/3}{x+2} dx.$$

Constants stay in front:

$$= \frac{7}{3}\ln|x-1| + \frac{2}{3}\ln|x+2| + C.$$

You can combine logs if desired:

$$\ln \left| (x-1)^{7/3}(x+2)^{2/3} \right| + C.$$

Both forms are acceptable unless directions say otherwise.

Always include absolute values and +C for indefinite integrals.

Definite integrals

If you’re evaluating something like

$$\int_0^2 \frac{3x+4}{(x-1)(x+2)} dx,$$

you:

1. Decompose.
2. Integrate.
3. Plug in upper minus lower.
4. No +C.

Be alert for vertical asymptotes inside the interval. If one exists, the integral may be improper, which changes the setup.

On AP free-response, algebra mistakes usually happen after the integration when plugging bounds into logs. Use parentheses carefully.

Visualizing what’s happening

Here’s the structure of what you’re doing algebraically in a concrete example:

Example of partial fraction decomposition

You’re rewriting one rational expression as a sum of simpler ones. The function hasn’t changed. Just its form.