AP Calculus BC

U-Substitution vs Partial Fractions

Q: What if the denominator does not factor?

Complete the square instead. A denominator like x^2 + 4x + 13 becomes (x+2)^2 + 9, which points to an \arctan answer rather than a decomposition.

Q: Can one integral need both methods?

Often. Decompose first, then substitute inside a piece such as \int \frac{1}{2x+5}\,dx, where u = 2x+5 supplies the factor of \frac{1}{2} that is easy to lose.

Use substitution when the integrand holds an inner function together with its own derivative as a factor. Use partial fractions when the integrand is a rational function whose denominator factors and no inner derivative appears, so you split it into simpler fractions and integrate each as a logarithm.

U-substitution

Use when: You can name an inner function $u$ and find $u'$ already sitting in the integrand as a factor, up to a constant multiple.

Partial fractions

Use when: The integrand is a rational function whose denominator factors into distinct linear pieces and no derivative of the denominator is present. This is CED topic 6.12, which is BC only.

Side by side

	U-substitution	Partial fractions
What it does	Collapses a composition into a standard form	Splits one fraction into a sum of simpler fractions
Look for	$f(g(x)) \cdot g'(x)$	$\frac{P(x)}{Q(x)}$ with $Q$ factorable and $\deg P < \deg Q$
Test case	$\int \frac{2x}{x^2 - 1}\,dx$	$\int \frac{1}{x^2 - 1}\,dx$
Answer	$\ln\lvert x^2 - 1 \rvert + C$	$\frac{1}{2}\ln\left\lvert \frac{x-1}{x+1} \right\rvert + C$
Common trap	Converting $x$ but leaving $dx$ in place of $du$	Skipping the long division when $\deg P \ge \deg Q$

Those two test cases differ by a single factor of $2x$ , and that factor decides the method. With it, $u = x^2 - 1$ gives $du = 2x\,dx$ and the integral becomes $\int \frac{du}{u}$ . Without it, no substitution helps, so factor the denominator as $(x-1)(x+1)$ and split the fraction instead.

Is the integrand rational? If not, partial fractions is off the table and substitution or parts is the route.
Is the derivative of the denominator present in the numerator, up to a constant? Substitute, and expect a logarithm.
Is the numerator degree at least the denominator degree? Long divide first, then decompose the remainder.
Does the denominator factor into distinct linear terms? Decompose, solve for the constants, and integrate term by term.

What the exam actually asks

The CED limits partial fractions to denominators that factor into distinct linear terms, so repeated factors and irreducible quadratics stay off the AP exam. Every piece then integrates to a logarithm, which is why these answers land as $\ln$ of a quotient.

Frequently asked questions

What if the denominator does not factor?

Complete the square instead. A denominator like $x^2 + 4x + 13$ becomes $(x+2)^2 + 9$ , which points to an $\arctan$ answer rather than a decomposition.

Can one integral need both methods?

Often. Decompose first, then substitute inside a piece such as $\int \frac{1}{2x+5}\,dx$ , where $u = 2x+5$ supplies the factor of $\frac{1}{2}$ that is easy to lose.

In the CED: Unit 6: Integration and Accumulation