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Linear
Partial Fractions

Master the technique of decomposing rational expressions into simpler fractions to unlock powerful integration strategies.

AP Calculus BC

What Are Partial Fractions?

Partial fraction decomposition is the process of breaking a complex rational expression into a sum of simpler fractions whose denominators are the factors of the original denominator.

Why Partial Fractions Matter

Many integrals involving rational functions are difficult or impossible to evaluate directly. Partial fractions transform them into sums of simple integrals we already know how to solve.

Hard
\(\displaystyle\int \frac{3x+5}{x^2+3x+2}\,dx\)
Easy
\(\displaystyle\int \frac{2}{x+1}\,dx + \int \frac{1}{x+2}\,dx\)

When Does It Apply?

1
Proper fraction: The degree of the numerator must be less than the degree of the denominator. If not, perform polynomial long division first.
2
Factorable denominator: The denominator can be factored into linear factors such as \((x - a)\), or irreducible quadratic factors.
3
Linear partial fractions apply when all denominator factors are linear (degree 1), either distinct or repeated.

The Decomposition Rule

Each \(A_i\) is a constant to be determined. The number of unknowns always equals the degree of the denominator.

Visualizing the Decomposition

\(\frac{P(x)}{(x-a)(x-b)(x-c)}\)
 =
\(\frac{A}{x - a}\)
 +
\(\frac{B}{x - b}\)
 +
\(\frac{C}{x - c}\)
 First factor Second factor Third factor

Distinct vs Repeated Factors

Distinct Linear Repeated Linear
Denominator \((x-1)(x+2)\) \((x-1)(x+2)^2\)
Setup \(\frac{A}{x-1}+\frac{B}{x+2}\) \(\frac{A}{x-1}+\frac{B}{x+2}+\frac{C}{(x+2)^2}\)
Unknowns 2 3
Key rule One term per factor One term per power

Step-by-Step Method

1
Check degree: Ensure \(\deg(P) < \deg(Q)\). If not, do polynomial long division first.
2
Factor the denominator completely into linear factors.
3
Write the decomposition with unknown constants \(A, B, C, \ldots\) in each numerator.
4
Clear the denominators by multiplying both sides by the original denominator.
5
Solve for constants using strategic substitution (plug in the roots) or by comparing coefficients.
6
Integrate each simple fraction separately.

Example 1: Distinct Factors

1
Set up: \(\frac{5x-3}{(x-1)(x+2)} = \frac{A}{x-1} + \frac{B}{x+2}\)
2
Clear denominators: \(5x - 3 = A(x+2) + B(x-1)\)
3
Substitution: Let \(x = 1\): \(2 = 3A \Rightarrow\) \(A = \frac{2}{3}\). Let \(x = -2\): \(-13 = -3B \Rightarrow\) \(B = \frac{13}{3}\).

The Substitution Trick

After clearing denominators, we get an identity that holds for all values of \(x\). Choosing the roots of the denominator factors eliminates terms and isolates one unknown at a time.

\(5x - 3 = A(x + 2) + B(x - 1)\)
Set \(x = 1\) (root of \(x - 1\))
Set \(x = -2\) (root of \(x + 2\))
\(2 = 3A \Rightarrow A = \tfrac{2}{3}\)
\(-13 = -3B \Rightarrow B = \tfrac{13}{3}\)
 Each root eliminates the other terms, leaving one unknown.

Example 2: Repeated Factor

1
Set up: \(\frac{2x+5}{(x+1)^2} = \frac{A}{x+1} + \frac{B}{(x+1)^2}\)
2
Clear denominators: \(2x + 5 = A(x+1) + B\)
3
Solve: Let \(x = -1\): \(3 = B\). Compare \(x\)-coefficients in \(2x+5 = Ax + (A+B)\): \(2=A\). So \(A=2\), \(B=3\).

Example 3: Three Factors

1
Setup: \(\frac{A}{x} + \frac{B}{x-1} + \frac{C}{x+1}\)
2
Clear: \(x^2 + 2x - 1 = A(x-1)(x+1) + Bx(x+1) + Cx(x-1)\)
3
Substitute: \(x=0\): \(-1 = -A \Rightarrow\) \(A = 1\). \(x=1\): \(2 = 2B \Rightarrow\) \(B = 1\). \(x=-1\): \(-2 = 2C \Rightarrow\) \(C = -1\).

Integrating Each Form

Simple
$$\int \frac{A}{x-a}\,dx = A\ln|x-a| + C$$
Repeated (\(n \ge 2\))
$$\int \frac{A}{(x-a)^n}\,dx = \frac{A}{(1-n)(x-a)^{n-1}} + C$$

Comparing Coefficients

When substitution does not give you enough equations (e.g., with repeated factors), expand both sides and match coefficients of like powers of \(x\).

Substitution
Best for distinct linear factors: fast and direct.
Coefficient Matching
Needed when repeated factors don't give enough equations.

Common Mistakes to Avoid

✕ Improper Fraction
Attempting decomposition when \(\deg(P) \ge \deg(Q)\). You must do long division first to reduce to a proper fraction.
✕ Missing Repeated Terms
For \((x-a)^3\), writing only \(\frac{A}{(x-a)^3}\) instead of all three terms: \(\frac{A}{x-a} + \frac{B}{(x-a)^2} + \frac{C}{(x-a)^3}\).
✕ Sign Errors
Forgetting the negative in factors like \((x - (-2)) = (x + 2)\). Be careful when substituting negative roots.

Summary Table

Scenario Decomposition Integral Result
\(\frac{A}{x - a}\) One simple fraction \(A\ln|x-a| + C\)
\(\frac{A}{(x-a)^n}, n \ge 2\) Term from repeated factor \(\frac{A}{(1-n)(x-a)^{n-1}} + C\)
Two distinct \(\frac{A}{x-a} + \frac{B}{x-b}\) Sum of two logs
Distinct + repeated Mixed terms Logs + power terms

Question 1 of 5

What is the correct partial fraction form for \(\frac{3x+1}{(x-2)(x+4)}\)?

Question 2 of 5

Given \(\frac{7x - 1}{(x-3)(x+1)} = \frac{A}{x-3} + \frac{B}{x+1}\), find \(A\).

Question 3 of 5

How many unknown constants are needed in the partial fraction decomposition of \(\frac{x+7}{x(x-1)^2(x+3)}\)?

Question 4 of 5

Evaluate \(\displaystyle\int \frac{4}{(x+1)(x-1)}\,dx\).

Question 5 of 5

Before applying partial fractions to \(\frac{x^3 + 2}{x^2 - 1}\), what must you do first?