How do I decide between u-substitution and integration by parts?

Try substitution first. If one factor is the derivative (up to a constant) of another factor's inside function, use u-substitution. If the integrand is a product of unrelated function types—like polynomial times exponential or polynomial times log—and no such derivative relationship exists, use integration by parts.

When do I need long division before partial fractions?

Whenever the numerator's degree is greater than or equal to the denominator's degree. Long division makes the rational function proper first; only then can partial fractions be applied to the leftover fraction. Skipping this check leads to an unsolvable decomposition.

What's the fastest way to recognize a trig-identity problem?

Look for even powers of sine or cosine (like ^2 x or ^4 x) or products such as x x that you cannot directly substitute. Rewrite using power-reduction or Pythagorean identities to convert them into standard integrable forms before integrating.

Can one integral require more than one technique?

Yes, frequently. You might do long division, then substitution, or substitution followed by integration by parts. On free-response questions, show each stage clearly, because setup earns partial credit even if a later computation slips.

AP-CALCBC-6.14

U6.14 Selecting Techniques for Antidifferentiation

Learn to quickly choose the right antidifferentiation technique on AP Calc BC—direct, u-substitution, parts, partial fractions, long division, or trig identities.

What you'll do in this lesson

A voice-first session with the Crimsora tutor on U6.14 Selecting Techniques for Antidifferentiation, then targeted practice and FRQs — with the tutor adapting to where you get stuck.

Start U6.14 Selecting Techniques for Antidifferentiation free

What this lesson covers

By this point in Unit 6 you know how to execute every major integration technique. Topic 6.14 tests something different: given an unfamiliar integral, can you recognize which method to use before wasting time? The AP exam rewards students who pattern-match fast, especially in the no-calculator sections where a wrong first choice can burn precious minutes.

This lesson builds a decision framework. You will learn diagnostic questions to ask about any integrand—Is it a standard form? Is there an inside function whose derivative also appears? Is it a product of unlike functions? Is it a rational function? Once you internalize these triggers, selecting a technique becomes nearly automatic.

A Decision Order That Works

When you meet an integral, run through techniques in order of speed, not alphabetically. The fastest wins.

First, ask whether the integrand is already a basic antiderivative you have memorized—powers,

e^x

\sin x

\frac{1}{x}

\frac{1}{1+x^2}

, and so on. If so, integrate directly.

Second, look for u-substitution: is there a composite function where the derivative of the inside factor is present (up to a constant)? This is the single most common trigger on the exam.

Third, if the integrand is a product of two different function types—polynomial times exponential, polynomial times log, exponential times trig—consider integration by parts.

Fourth, if the integrand is a rational function

\frac{P(x)}{Q(x)}

, check the degrees. If

\deg P \ge \deg Q

, do long division first. If

\deg P < \deg Q

and the denominator factors, use partial fractions.

Fifth, if you see even powers of sine and cosine, or products you cannot substitute, apply a trig identity to rewrite before integrating.

Trigger	Technique
Memorized form	Direct antiderivative
Inside function + its derivative	u-substitution
Product of unlike functions	Integration by parts
Rational, $\deg P \ge \deg Q$	Long division
Rational, factorable denominator	Partial fractions
Even powers / stubborn trig	Trig identity

Spotting u-Substitution vs. Integration by Parts

These two are the most confused. The key distinction: u-substitution undoes the chain rule, while integration by parts undoes the product rule.

Use u-substitution when one factor is (a constant multiple of) the derivative of an inside function. Consider

\int 2x\,e^{x^2}\,dx

. Here

\frac{d}{dx}(x^2)=2x

appears explicitly, so let

u=x^2

and the integral collapses to

\int e^u\,du

.

Use integration by parts when the two factors are unrelated by differentiation—neither is the derivative of the other's inside. Consider

\int x e^x\,dx

. There is no inside function whose derivative is present; instead you have a polynomial times an exponential. Apply

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

with

u=x

dv=e^x\,dx

.

A reliable test: try substitution first. If setting

u

equal to the inside function does not eliminate the leftover factor, substitution fails and parts is likely correct.

The LIATE guideline (Logarithmic, Inverse trig, Algebraic, Trig, Exponential) tells you which factor to call

u

in parts: choose the one appearing earliest in the list. So in

\int x\ln x\,dx

, the log wins and

u=\ln x

.

A common misconception is reaching for parts on

\int x\sqrt{x^2+1}\,dx

. That is a substitution problem—

u=x^2+1

—because

x

is essentially the derivative of the inside.

Rational Functions: Division, Factoring, and Completing the Square

Rational integrands demand a quick structural check before any calculation.

Step one is comparing degrees. If the numerator's degree is greater than or equal to the denominator's, you must perform polynomial long division first. For example,

\int \frac{x^2}{x+1}\,dx

has numerator degree 2 and denominator degree 1, so divide to get

x-1+\frac{1}{x+1}

, then integrate term by term.

Step two, once the fraction is proper, examine the denominator. If it factors into distinct linear (or repeated linear) factors, use partial fractions: write

\frac{1}{(x-1)(x+2)} = \frac{A}{x-1}+\frac{B}{x+2}

and solve for the constants.

Step three, if the denominator is an irreducible quadratic, consider completing the square to reach an arctangent or logarithm form. For instance

\int \frac{dx}{x^2+4x+5}

becomes

\int \frac{dx}{(x+2)^2+1}

, which is

\arctan(x+2)+C

Denominator type	Move
Degree of top too big	Long division
Factorable	Partial fractions
Irreducible quadratic	Complete the square

On the exam, do not launch into partial fractions before checking degrees—a common error that leads to an impossible decomposition.

How the Exam Tests Technique Selection

AP questions rarely say "use integration by parts." Instead they hand you an integral and expect the right choice to be part of your work. In the multiple-choice section, recognizing the method instantly is worth more than raw computation speed, because the answer choices are close.

Watch for disguised basic forms. The integrand

\frac{2x}{x^2+1}

looks like it needs partial fractions, but the numerator is the derivative of the denominator, so a quick u-substitution gives

\ln|x^2+1|+C

.

Watch for trig identity setups. You cannot directly antidifferentiate

\sin^2 x

; rewrite it using

\sin^2 x = \frac{1-\cos 2x}{2}

. Similarly

\tan^2 x = \sec^2 x - 1

turns a hard integral into two easy ones.

Finally, remember that techniques combine. A single integral may need long division, then substitution. A parts problem may require a substitution inside. In free-response, show each stage clearly; partial credit follows correct setup even if arithmetic slips.

The biggest time-saver is a two-second scan before writing anything: identify the integrand's structure, name the technique, then execute.

Key terms

Direct antiderivative.: Integrating a function that already matches a memorized rule, such as $\int e^x\,dx = e^x + C$ , with no rewriting needed.
u-substitution.: A method that reverses the chain rule by letting $u$ equal an inside function whose derivative also appears in the integrand.
Integration by parts.: A method reversing the product rule, using $\int u\,dv = uv - \int v\,du$ , best for products of unlike function types.
Partial fractions.: Decomposing a proper rational function into a sum of simpler fractions whose denominators are the factors of the original denominator.
Long division of polynomials.: Rewriting an improper rational function (numerator degree ≥ denominator degree) as a polynomial plus a proper remainder fraction before integrating.
Completing the square.: Rewriting an irreducible quadratic denominator in the form $(x+a)^2+b$ to reach an arctangent or logarithmic antiderivative.
LIATE.: A guideline (Logarithmic, Inverse trig, Algebraic, Trig, Exponential) for choosing which factor to assign as $u$ in integration by parts.
Trig identity rewrite.: Using identities like $\sin^2 x=\frac{1-\cos 2x}{2}$ or $\tan^2 x=\sec^2 x-1$ to convert an unintegrable trig expression into standard forms.

Worked example

Evaluate

\int \frac{x^3}{x^2+1}\,dx

First, scan the structure. This is a rational function, so compare degrees: the numerator has degree 3 and the denominator degree 2. Because

3 \ge 2

, the fraction is improper, so long division comes first—not partial fractions.

Divide

x^3

x^2+1

. Since

x^3 = x(x^2+1) - x

, we get

\frac{x^3}{x^2+1} = x - \frac{x}{x^2+1}

.

Now integrate term by term. The first piece is direct:

\int x\,dx = \frac{x^2}{2}

.

For the second piece,

\int \frac{x}{x^2+1}\,dx

, notice the numerator

x

is (up to a factor) the derivative of the denominator

x^2+1

. This is a u-substitution trigger. Let

u = x^2+1

, so

du = 2x\,dx

, meaning

x\,dx = \frac{1}{2}du

. Then

\int \frac{x}{x^2+1}\,dx = \frac{1}{2}\int \frac{du}{u} = \frac{1}{2}\ln|u| = \frac{1}{2}\ln(x^2+1)

.

Combine:

\int \frac{x^3}{x^2+1}\,dx = \frac{x^2}{2} - \frac{1}{2}\ln(x^2+1) + C

.

This problem shows technique layering—long division sets up a substitution—which is exactly what the exam expects you to recognize.

Practice questions

Which technique most efficiently evaluates

\int x^2 \ln x\,dx

u-substitution with $u=x^2$
Integration by parts with $u=\ln x$
Partial fractions
Long division

Answer: Integration by parts with $u=\ln x$

The integrand is a product of an algebraic factor

x^2

and a logarithmic factor

\ln x

—unlike function types with no inside-derivative relationship, so u-substitution fails. This signals integration by parts. By LIATE, the logarithm is chosen as

u

, giving

u=\ln x

dv=x^2\,dx

. Then

du=\frac{1}{x}dx

and

v=\frac{x^3}{3}

, producing a manageable remaining integral

\int \frac{x^3}{3}\cdot\frac{1}{x}\,dx

For

\int \frac{5}{(x-1)(x+4)}\,dx

, identify and briefly justify the most efficient technique, then set up (do not fully solve) the first step.

Answer: Partial fractions; write $\frac{5}{(x-1)(x+4)} = \frac{A}{x-1} + \frac{B}{x+4}$ .

The integrand is a proper rational function (numerator degree 0, denominator degree 2) whose denominator is already factored into distinct linear factors. That is the textbook trigger for partial fractions—no long division is needed because the fraction is already proper. Solving

5 = A(x+4)+B(x-1)

yields the constants, after which each term integrates to a natural log.

Explain why

\int 3x^2\cos(x^3)\,dx

should be solved by substitution rather than by parts.

Answer: Because $3x^2$ is exactly the derivative of the inside function $x^3$ , a substitution $u=x^3$ collapses the integral to $\int \cos u\,du$ .

The presence of a composite function

\cos(x^3)

together with

3x^2

—the derivative of its inside—signals the chain rule is at work, which u-substitution reverses. Integration by parts would be circular here because the factors are related by differentiation, not independent. With

u=x^3

du=3x^2\,dx

, so the integral becomes

\int\cos u\,du = \sin(x^3)+C

FAQ

How do I decide between u-substitution and integration by parts?: Try substitution first. If one factor is the derivative (up to a constant) of another factor's inside function, use u-substitution. If the integrand is a product of unrelated function types—like polynomial times exponential or polynomial times log—and no such derivative relationship exists, use integration by parts.
When do I need long division before partial fractions?: Whenever the numerator's degree is greater than or equal to the denominator's degree. Long division makes the rational function proper first; only then can partial fractions be applied to the leftover fraction. Skipping this check leads to an unsolvable decomposition.
What's the fastest way to recognize a trig-identity problem?: Look for even powers of sine or cosine (like $\sin^2 x$ or $\cos^4 x$ ) or products such as $\sin x\cos x$ that you cannot directly substitute. Rewrite using power-reduction or Pythagorean identities to convert them into standard integrable forms before integrating.
Can one integral require more than one technique?: Yes, frequently. You might do long division, then substitution, or substitution followed by integration by parts. On free-response questions, show each stage clearly, because setup earns partial credit even if a later computation slips.

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