Integration using partial fractions (BC only) — AP Calculus BC
AP Calculus BC · AP Calculus BC CED Unit 6: Integration and Accumulation of Change · 14 min read
1. Proper vs Improper Rational Functions and Long Division★★☆☆☆⏱ 3 min
Partial fraction decomposition can only be applied directly to proper rational functions. If the function is improper ($\deg(P) \geq \deg(Q)$), you must first rewrite it as the sum of a polynomial and a new proper rational function using polynomial long division. For AP Calculus BC, you will only encounter linear factors (distinct or repeated) in the denominator; irreducible quadratic factors are not tested.
2. Partial Fractions with Distinct Linear Factors★★★☆☆⏱ 4 min
The most common case on the exam is a denominator that factors into distinct (non-repeating) linear factors. For this case, the decomposition follows the rule:
To solve for constants, multiply both sides by $Q(x)$ to eliminate denominators. The fastest method is root substitution: substitute the root of each linear factor into the resulting equation to solve for $A_i$ directly, since all other terms become zero. Integrate term-by-term once you have all constants.
3. Partial Fractions with Repeated Linear Factors★★★★☆⏱ 4 min
If a linear factor is repeated $k$ times in the denominator (i.e., $(ax+b)^k$ for $k \geq 2$), you must add a separate term for every power of the factor from 1 up to $k$. For example, if the denominator is $x(x-1)^2$, the decomposition is $\frac{A}{x} + \frac{B}{x-1} + \frac{C}{(x-1)^2}$.
When integrating, terms with power 1 still integrate to a logarithm, while terms with power $k \geq 2$ integrate using the power rule: $\int (ax+b)^{-k} dx = \frac{(ax+b)^{1-k}}{a(1-k)} + C$.
4. AP-Style Worked Practice Problems★★★★☆⏱ 3 min
Common Pitfalls
Why: Students rush to start partial fractions without checking degrees, forgetting decomposition only works for proper rationals.
Why: Students confuse distinct and repeated factor rules, forgetting each power from 1 up to the exponent needs its own term.
Why: Students remember the antiderivative of $\frac{1}{x}$ is $\ln x$ from early lessons, forgetting $\ln x$ is only defined for positive $x$, but $\frac{1}{x}$ is defined for negative $x$ too.
Why: Students rush substitution and do not check their work.
Why: Students rush the factoring step, which undermines the entire problem.