AP Calculus AB and BC

U-Sub vs Integration by Parts vs Partial Fractions

Match the integrand's structure to the technique: a factor that is the derivative of an inside function means u-substitution; a product of unrelated functions or a lone log/inverse-trig function means integration by parts (BC only); a rational function with a factorable denominator means partial fractions (BC only).

Read the integrand before you reach for a rule

Topic 6.14 does not add a new integration technique. It tests one skill: looking at an unfamiliar integral and naming the right tool before you compute. On the AP exam that recognition step is where most points are won or lost, because the wrong technique burns the whole problem. The habit to build is simple. Read the integrand, classify its structure, then commit to a method.

Three questions sort almost every integral you will see. Is one factor the derivative of something inside the expression? Is the integrand a product of two functions from unrelated families? Is it a rational function, one polynomial divided by another? Those answers point to u-substitution, integration by parts, and partial fractions in that order. u-substitution (Topic 6.9) is on both AB and BC. Integration by parts (Topic 6.11) and partial fractions (Topic 6.12) are BC only, so AB students never need them. Work through the questions in that same order every time, because the earlier a technique sits, the less algebra it usually costs you.

What you see in the integrand	Technique	CED topic	Course
A factor is the derivative of an inside function (up to a constant multiple)	u-substitution	6.9	AB and BC
Product of two unrelated types (polynomial times $e^x$ , $\sin x$ , or $\ln x$ ), or a lone $\ln x$ or inverse-trig function	Integration by parts	6.11	BC only
Rational function whose denominator factors into distinct linear pieces, numerator degree below denominator degree	Partial fractions	6.12	BC only
Rational function with numerator degree at least the denominator degree	Long division first, then re-decide	6.10	AB and BC
Irreducible quadratic in the denominator	Complete the square, expect an arctan	6.10	AB and BC

Test for u-substitution first

u-substitution is the fastest technique and it often turns a frightening integral into a basic form. Always check for it before you commit to parts or partial fractions. Many integrals that look like harder problems collapse under one substitution.

When it is u-substitution (Topic 6.9)

u-substitution reverses the chain rule. It works when the integrand contains a composite function and the derivative of the inner function is also present as a factor. A constant multiple is fine, because you can adjust with a coefficient. The signal you are hunting for is a pairing: an inside expression and (a multiple of) its derivative both sitting in the same integral.

A composite like $e^{x^2}$ , $\sin(3x)$ , or $(x^2+1)^5$ multiplied by a factor that matches the inside function's derivative.
A fraction whose numerator is the derivative of its denominator: that is the $\int \frac{u'}{u}\,dx = \ln|u| + C$ pattern.
A power of a function multiplied by that function's derivative, such as $\int (\ln x)^2 \cdot \frac{1}{x}\,dx$ .

The method: let $u$ be the inner function, compute $du$ , and solve for the exact piece you need to replace. Rewrite the whole integral in terms of $u$ with no $x$ remaining. If any $x$ survives, your $u$ was the wrong choice. Integrate, then substitute back. For a definite integral, Topic 6.9 is explicit that you either convert the limits to $u$ -values or switch fully back to $x$ before evaluating; you cannot plug $x$ -limits into a $u$ -expression.

When it is integration by parts (Topic 6.11, BC only)

BC only

Integration by parts is Topic 6.11 and appears only on the BC exam. AB students can skip this section.

Integration by parts reverses the product rule. Reach for it when the integrand is a product of two functions from different families and neither one is the derivative of the other's inside, so substitution has nothing to cancel. That absence of a derivative pairing is the tell. The formula is:

\int u\,dv = uv - \int v\,du

The whole game is choosing $u$ and $dv$ well, and the mnemonic LIATE ranks the choice for $u$ : Logarithmic, Inverse trig, Algebraic (polynomials), Trigonometric, Exponential. Whatever appears earliest on that list becomes $u$ , the part you differentiate; everything else becomes $dv$ , the part you integrate. The point of the ranking is to make $u$ get simpler when differentiated and to leave a $dv$ you can actually integrate.

$\int x e^x\,dx$ : polynomial times exponential, so let $u = x$ .
$\int x \cos x\,dx$ : polynomial times trig, so let $u = x$ .
$\int \ln x\,dx$ : a lone logarithm, so let $u = \ln x$ and $dv = dx$ .
$\int x^2 e^x\,dx$ : expect to apply parts twice, since each pass lowers the polynomial's degree by one.

When it is partial fractions (Topic 6.12, BC only)

BC only

Linear partial fractions is Topic 6.12 and appears only on the BC exam. The course covers distinct linear, nonrepeating factors, so you will not see repeated factors or irreducible-quadratic decompositions.

Partial fractions breaks one complicated rational function into a sum of simple ones you already know how to integrate. Use it when the integrand is a rational function, the denominator factors into distinct linear factors, and the numerator's degree is strictly below the denominator's. Each simple piece integrates to a natural logarithm, which is why the whole technique reduces a hard-looking fraction to a sum of logs.

The method: factor the denominator, then write the fraction as a sum of unknown constants over each linear factor, for example $\frac{1}{(x-1)(x+2)} = \frac{A}{x-1} + \frac{B}{x+2}$ . Clear the denominators, then solve for the constants. Substituting the roots one at a time is the fastest way, because each root kills every term but one and hands you a constant directly. Once you have the constants, you are just integrating a short list of $\frac{1}{x-r}$ terms, each of which becomes $\ln|x-r|$ .

Two checks before you decompose

If the numerator degree is at least the denominator degree, do long division first (Topic 6.10). If the numerator is already the derivative of the denominator, the problem is a plain u-substitution, not partial fractions.

Common traps and how to avoid them

Most lost points on selection problems come from misreading the structure, not from bad algebra. Watch for these.

Skipping the u-substitution check. Integrals that look like parts or partial fractions often collapse under a substitution. $\int \frac{x}{x^2+4}\,dx$ looks like a partial-fractions candidate but is really u-substitution, because the numerator is a constant multiple of the denominator's derivative.
Forgetting to change the limits (or to switch back to $x$ ) on a definite u-substitution. Topic 6.9 requires the limits to match the variable you are integrating in.
Choosing $u$ and $dv$ backwards in integration by parts, which makes $\int v\,du$ harder than the original integral. Let LIATE decide.
Trying partial fractions when the numerator degree is too high. Divide first, then decompose the remainder.
Dropping the constant of integration $+C$ on an indefinite integral.

Worked examples

Worked example

Recognizing u-substitution in disguise

Evaluate $\int \frac{x}{x^2+4}\,dx$ .

Classify first. The integrand is a rational function, so partial fractions is tempting. But check the u-substitution signal: is the numerator a multiple of the derivative of the denominator? The denominator is $x^2+4$ , and $\frac{d}{dx}(x^2+4) = 2x$ . The numerator $x$ is exactly half of that, so this is u-substitution, not partial fractions.
Set $u = x^2+4$ . Then $du = 2x\,dx$ , so $x\,dx = \frac{1}{2}\,du$ .
Rewrite the integral entirely in $u$ , leaving no $x$ behind: $\int \frac{x}{x^2+4}\,dx = \int \frac{1}{u}\cdot \frac{1}{2}\,du = \frac{1}{2}\int \frac{1}{u}\,du$ .
Integrate the basic form: $\frac{1}{2}\int \frac{1}{u}\,du = \frac{1}{2}\ln|u| + C$ .
Substitute back. Since $x^2+4 > 0$ for every $x$ , the absolute value is optional: $\frac{1}{2}\ln(x^2+4) + C$ .

$\frac{1}{2}\ln(x^2+4) + C$

Worked example

Integration by parts: polynomial times exponential (BC)

Evaluate $\int x e^{2x}\,dx$ .

Classify. The integrand is a product of an algebraic factor $x$ and an exponential $e^{2x}$ . Neither is the derivative of the other's inside, so a substitution has nothing to cancel. That is the integration-by-parts signal (BC only, Topic 6.11).
Apply LIATE. Algebraic outranks Exponential, so let $u = x$ (the part you differentiate) and $dv = e^{2x}\,dx$ (the part you integrate).
Compute the pieces: $du = dx$ , and $v = \int e^{2x}\,dx = \frac{1}{2}e^{2x}$ .
Substitute into $\int u\,dv = uv - \int v\,du$ : $\int x e^{2x}\,dx = x\cdot \frac{1}{2}e^{2x} - \int \frac{1}{2}e^{2x}\,dx$ .
The remaining integral is basic: $\int \frac{1}{2}e^{2x}\,dx = \frac{1}{2}\cdot \frac{1}{2}e^{2x} = \frac{1}{4}e^{2x}$ .
Combine and add the constant: $\frac{1}{2}x e^{2x} - \frac{1}{4}e^{2x} + C$ .

$\frac{1}{2}x e^{2x} - \frac{1}{4}e^{2x} + C$

Worked example

Partial fractions with distinct linear factors (BC)

Evaluate $\int \frac{1}{x^2+x-2}\,dx$ .

Classify. This is a rational function with a constant numerator, so the numerator degree (0) is below the denominator degree (2). Factor the denominator: $x^2+x-2 = (x+2)(x-1)$ , two distinct linear factors. The numerator is not the derivative of the denominator, since $\frac{d}{dx}(x^2+x-2) = 2x+1$ is not a constant, so this is partial fractions (BC only, Topic 6.12), not u-substitution.
Set up the decomposition: $\frac{1}{(x+2)(x-1)} = \frac{A}{x+2} + \frac{B}{x-1}$ .
Clear denominators by multiplying both sides by $(x+2)(x-1)$ : $1 = A(x-1) + B(x+2)$ .
Solve for the constants using the roots. Let $x = 1$ : $1 = A(0) + B(3)$ , so $B = \frac{1}{3}$ . Let $x = -2$ : $1 = A(-3) + B(0)$ , so $A = -\frac{1}{3}$ .
Rewrite and integrate each piece: $\int \left(\frac{-1/3}{x+2} + \frac{1/3}{x-1}\right)\,dx = -\frac{1}{3}\ln|x+2| + \frac{1}{3}\ln|x-1| + C$ .
Combine the logs into one term: $\frac{1}{3}\ln\left|\frac{x-1}{x+2}\right| + C$ .

$\frac{1}{3}\ln\left|\frac{x-1}{x+2}\right| + C$

Frequently asked questions

Is integration by parts on the AP Calculus AB exam?

No. Integration by parts (Topic 6.11) and partial fractions (Topic 6.12) are BC only. AB students handle antidifferentiation with u-substitution (Topic 6.9), the basic antiderivative rules, long division, and completing the square (Topic 6.10).

How do I know whether to use u-substitution or integration by parts?

Check for a derivative pairing first. If one factor is a constant multiple of the derivative of an inside function, use u-substitution. If the integrand is a product of two unrelated functions with no such pairing, use integration by parts. When in doubt, always try the substitution before parts.

What is LIATE and when do I use it?

LIATE stands for Logarithmic, Inverse trig, Algebraic, Trigonometric, Exponential. In integration by parts, it tells you which factor to call u: pick the one earliest on the list to differentiate, and let the rest be dv. It only matters on BC, since parts is a BC topic.

Can you use partial fractions when the denominator does not factor into linear pieces?

Not on the AP exam. Topic 6.12 covers only distinct linear, nonrepeating factors. An irreducible quadratic denominator is handled by completing the square (Topic 6.10), which usually leads to an arctan rather than a log.