Unit 6Integration by Substitution

Concept 4.87Built for May 2027 AP Exam

Integration by Substitution

Reviewed by ACEpath AP Math Editorial Team

Last Updated: September 2026

30-second summary

Substitution runs the chain rule in reverse. When an integrand contains an inner function and something close to its derivative, name the inner function $u$ , write $du$ , and the integral collapses into one you already know. For an indefinite integral, finish by returning to $x$ . For a definite integral, change the limits into $u$ -values and never go back, because mixing $x$ -limits with a $u$ -integrand is the error that ruins otherwise correct work. The whole skill is pattern recognition: the derivative of the inside has to be present up to a constant factor, and that constant is the only thing you are allowed to adjust.

On the AP Exam

AP Calculus AB tests this skill through the representations and reasoning described below.

What you’ll see: Seldom as a free-response part of its own. It appears inside differential-equation solutions and inside multiple-choice items, usually as $\int x\,e^{x^{2}}dx$ or $\int \frac{2x}{x^{2}+1}\,dx$ , and often as one step of a longer computation.
What you’ll do: Pick $u$ as the inner function, compute $du$ , match the remaining factor, and then either back-substitute or convert the limits.
Memory hook: Name the inside, then follow it down. Whatever sits inside the parentheses or the exponent becomes $u$ , and everything else must become $du$ .
Use it on the exam: For MCQs: Differentiate the choices rather than integrating them. A candidate answer that differentiates back to the integrand is correct, and the check bypasses the whole substitution. For FRQs: Write $u$ and $du$ on separate lines before rewriting the integral. When the integral is definite, write the new limits immediately, in the same line, so they cannot be forgotten.

Concept Review

The Chain Rule, Reversed

Figure 1 — Substitution changes the variable and the interval together. The region on the $x$ -axis maps to a region on the $u$ -axis, and the limits must travel with it.

The chain rule says $\frac{d}{dx}F\left(g(x)\right) = F'\left(g(x)\right)g'(x)$ . Integrating both sides gives the substitution rule:

$\int f\left(g(x)\right)g'(x)\,dx = F\left(g(x)\right) + C, \qquad \text{with } u = g(x),\ du = g'(x)\,dx$

In practice you never write $F$ . You write $u$ , you write $du$ , and the integral becomes $\int f(u)\,du$ .

Substitution changes the variable and the interval together. The region on the $x$-axis maps to a region on the $u$-axis, and the limits must travel with it.

Choosing $u$

Integrand	Natural choice of $u$	Why
$x\,e^{x^{2}}$	$u = x^{2}$	$du = 2x\,dx$ , and the $x\,dx$ is present up to the $2$
$\frac{2x}{x^{2}+1}$	$u = x^{2}+1$	$du = 2x\,dx$ matches exactly
$\cos(3x)\,dx$	$u = 3x$	$du = 3\,dx$
$\sin^{4}x\cos x\,dx$	$u = \sin x$	$du = \cos x\,dx$
$x\sqrt{x^{2}+4}\,dx$	$u = x^{2}+4$	$du = 2x\,dx$ , present up to a factor

The pattern in every row is the same: the derivative of the inside appears somewhere in the integrand, possibly multiplied by a constant you can move.

Adjusting a Constant Is Legal; Adjusting a Variable Is Not

If $du = 2x\,dx$ and the integrand contains only $x\,dx$ , you may write $x\,dx = \frac{1}{2}du$ and carry the $\frac{1}{2}$ outside. That is multiplication by a constant, and constants pass through an integral. What you may not do is invent a missing factor of $x$ , because $x$ is not a constant and it changes with the variable.

$\int x\,e^{x^{2}}dx = \frac{1}{2}\int e^{u}\,du = \frac{1}{2}e^{u} + C = \frac{1}{2}e^{x^{2}} + C$

Definite Integrals: Change the Limits

There are two correct routes and one wrong one. The correct routes are to convert the limits into $u$ -values before integrating, or to back-substitute and then evaluate at the original $x$ -limits. The wrong route is to integrate in $u$ and then substitute $x$ -limits into a $u$ -expression.

$\int_0^{1} x\,e^{x^{2}}dx = \frac{1}{2}\int_{0}^{1} e^{u}\,du = \frac{1}{2}\left[e^{u}\right]_0^{1} = \frac{1}{2}(e - 1)$

The new limits come from the substitution itself: $u = x^{2}$ sends $x = 0$ to $u = 0$ and $x = 1$ to $u = 1$ . When the substitution is not monotone on the interval, the converted limits can collapse or reverse, and both are correct behaviour rather than errors.

When Substitution Fails

Substitution needs the inner derivative to be present. $\int e^{x^{2}}dx$ has no elementary antiderivative, and no choice of $u$ fixes that, because the missing factor $x$ cannot be manufactured. Recognizing the failure is part of the skill: on the exam such integrands appear only inside definite integrals where a calculator or a table supplies the value.

Common Misconception Check

Myth 1: “Substitution works on any composite integrand.” Truth: It works when the inner derivative is present. Without that factor the method has nothing to absorb.
Myth 2: “After substituting, the old limits still apply.” Truth: The limits are $x$ -values and the integrand is now a $u$ -expression. Both must speak the same variable.
Myth 3: “You must always convert the limits.” Truth: Back-substituting and using the original limits is equally valid, and often safer when the substitution is awkward to invert.

Topic-Specific Study Method

Name the Inside, Then Follow It

Borrowed idea: Find the inside. Name it $u$ . The inner function is what makes the integrand composite, and it sits either in the exponent or in the parentheses. Then ask what $du$ has to be. If the answer matches what is left over, the rest is one line of work. If it does not match, the method is wrong for this integrand, and no amount of rewriting will make the missing factor appear.

Exam disguise	What you see on the page	What your brain must translate	Required execution
Power of a trig function	$\sin^{4}x\cos x$	$u = \sin x$ , $du = \cos x\,dx$	absorb the cosine
Exponent with a square	$x\,e^{x^{2}}$	$du = 2x\,dx$ , so a $\frac{1}{2}$ appears	adjust the constant only
Fraction with a quadratic below	$\frac{2x}{x^{2}+1}$	$u$ is the denominator	gives a logarithm
Root of a quadratic	$x\sqrt{x^{2}+4}$	$u = x^{2}+4$	power rule on $u^{1/2}$
Definite integral with a composite	limits plus an inner function	the limits must change too	convert before integrating

① Write $u$ and $du$ on their own lines. The two lines are the method, and skipping them is why students lose track of which variable is live.

② Convert the limits at the same moment you convert the integrand. A definite integral in $u$ with $x$ -limits is the single most common structural error in the topic.

③ Differentiate the final answer. For an indefinite integral this returns the original integrand, and for a definite integral it is not needed, since the numeric value can be sanity-checked by sign and size.

Training & Error Diagnosis

Core Archetype (Simplest Form): Find $\int 2x\left(x^{2}+3\right)^{5}dx$ .

Standard Execution:

Choose. $u = x^{2}+3$ , so $du = 2x\,dx$ .

Rewrite. The integral becomes $\int u^{5}\,du$ .

Integrate. $\frac{u^{6}}{6} + C$ .

Return. $\frac{\left(x^{2}+3\right)^{6}}{6} + C$ .

Check. Differentiating gives $\left(x^{2}+3\right)^{5}\cdot 2x$ , which matches.

Exam Variation (Context / Parameter Shift): Evaluate $\int_1^{2} \frac{3x^{2}}{x^{3}+1}\,dx$ .

Solution: Let $u = x^{3}+1$ , so $du = 3x^{2}dx$ and the integrand is exactly $\frac{du}{u}$ . The limits convert: $x = 1$ gives $u = 2$ , and $x = 2$ gives $u = 9$ . So the integral is $\int_2^{9}\frac{du}{u} = \left[\ln u\right]_2^{9} = \ln 9 - \ln 2 = \ln\frac{9}{2}$ . Note the two decisions that carry the points: the limits moved with the substitution, and the absolute value in the logarithm can be dropped because both limits are positive.

Common Traps & Failure Signals:

Limits left in $x$ (Signal: an $x$ -value substituted into a $u$ -expression): convert the limits or back-substitute, never mix.
Missing $du$ factor (Signal: an $x$ left over after substituting): the derivative of the inside must be fully absorbed.
Invented factor (Signal: multiplying by $x$ to make the match work): only constants may be adjusted.

6D Cognitive Error Diagnosis (Where Your Thinking Broke):

Inside Unnamed: Write $u$ before rewriting.
Variable Mixing: You converted the integrand and kept the old limits, which is a bookkeeping failure rather than a calculus one, and it produces a wrong number with no visible symptom because every individual step looks correct.
Factor Invented: You supplied the missing $x$ to make the substitution fit, which works only when the missing factor is a constant, and the resulting answer differentiates back to something that is not the original integrand at all.

Past 10-Year AP Exam Appearance

Testing Frequency: MCQ Only; No released free-response part since 2017 requires u-substitution as a scored step; every appearance is in the multiple-choice section or inside a differential-equation solution
10-Year Total Count: 0 of 9 Exams · 0 FRQ Parts; The nine free-response sets contain no part whose rubric mentions substitution
FRQ Years Appeared: None in Released FRQs ---

Practice Examples

MCQ 1

The Straightforward Case

$\int 2x\left(x^{2}+5\right)^{3}dx =$

MCQ 2

Changing the Limits

$\int_0^{1} 3x^{2}\left(x^{3}+1\right)^{2}dx =$

MCQ 3

Recognising a Logarithm

$\int \frac{4x}{x^{2}+9}\,dx =$

FRQ 4

Substitution in Both Directions

(a) Find $\int x^{2}\sqrt{x^{3}+1}\,dx$ .

(b) Evaluate $\int_0^{2} \frac{x}{x^{2}+1}\,dx$ .

(c) A student writes $\int_0^{2} \frac{x}{x^{2}+1}\,dx = \left[\frac{1}{2}\ln\left(x^{2}+1\right)\right]_0^{2}$ and then claims the substitution was never needed. Explain why the method is sound even though the antiderivative was written directly in $x$ .

Cheatsheet

Core Reference

	Action	Failure if skipped
1	name $u$ as the inner function	nothing collapses
2	compute $du$ and solve for the leftover	an $x$ survives
3	adjust constants only	the answer differentiates back wrong
4	convert the limits (definite)	$x$ -limits on a $u$ -integrand
5	back-substitute (indefinite)	the answer is in the wrong variable

Sentences That Score

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Unit 6Integration by Substitution

Concept 4.87Built for May 2027 AP Exam

Integration by Substitution

Reviewed by ACEpath AP Math Editorial Team

Last Updated: September 2026

30-second summary

On the AP Exam

AP Calculus AB tests this skill through the representations and reasoning described below.

What you’ll see: Seldom as a free-response part of its own. It appears inside differential-equation solutions and inside multiple-choice items, usually as $\int x\,e^{x^{2}}dx$ or $\int \frac{2x}{x^{2}+1}\,dx$ , and often as one step of a longer computation.
What you’ll do: Pick $u$ as the inner function, compute $du$ , match the remaining factor, and then either back-substitute or convert the limits.
Memory hook: Name the inside, then follow it down. Whatever sits inside the parentheses or the exponent becomes $u$ , and everything else must become $du$ .
Use it on the exam: For MCQs: Differentiate the choices rather than integrating them. A candidate answer that differentiates back to the integrand is correct, and the check bypasses the whole substitution. For FRQs: Write $u$ and $du$ on separate lines before rewriting the integral. When the integral is definite, write the new limits immediately, in the same line, so they cannot be forgotten.

Concept Review

The Chain Rule, Reversed

Figure 1 — Substitution changes the variable and the interval together. The region on the $x$ -axis maps to a region on the $u$ -axis, and the limits must travel with it.

The chain rule says $\frac{d}{dx}F\left(g(x)\right) = F'\left(g(x)\right)g'(x)$ . Integrating both sides gives the substitution rule:

$\int f\left(g(x)\right)g'(x)\,dx = F\left(g(x)\right) + C, \qquad \text{with } u = g(x),\ du = g'(x)\,dx$

In practice you never write $F$ . You write $u$ , you write $du$ , and the integral becomes $\int f(u)\,du$ .

Substitution changes the variable and the interval together. The region on the $x$-axis maps to a region on the $u$-axis, and the limits must travel with it.

Choosing $u$

Integrand	Natural choice of $u$	Why
$x\,e^{x^{2}}$	$u = x^{2}$	$du = 2x\,dx$ , and the $x\,dx$ is present up to the $2$
$\frac{2x}{x^{2}+1}$	$u = x^{2}+1$	$du = 2x\,dx$ matches exactly
$\cos(3x)\,dx$	$u = 3x$	$du = 3\,dx$
$\sin^{4}x\cos x\,dx$	$u = \sin x$	$du = \cos x\,dx$
$x\sqrt{x^{2}+4}\,dx$	$u = x^{2}+4$	$du = 2x\,dx$ , present up to a factor

The pattern in every row is the same: the derivative of the inside appears somewhere in the integrand, possibly multiplied by a constant you can move.

Adjusting a Constant Is Legal; Adjusting a Variable Is Not

$\int x\,e^{x^{2}}dx = \frac{1}{2}\int e^{u}\,du = \frac{1}{2}e^{u} + C = \frac{1}{2}e^{x^{2}} + C$

Definite Integrals: Change the Limits

$\int_0^{1} x\,e^{x^{2}}dx = \frac{1}{2}\int_{0}^{1} e^{u}\,du = \frac{1}{2}\left[e^{u}\right]_0^{1} = \frac{1}{2}(e - 1)$

When Substitution Fails

Common Misconception Check

Myth 1: “Substitution works on any composite integrand.” Truth: It works when the inner derivative is present. Without that factor the method has nothing to absorb.
Myth 2: “After substituting, the old limits still apply.” Truth: The limits are $x$ -values and the integrand is now a $u$ -expression. Both must speak the same variable.
Myth 3: “You must always convert the limits.” Truth: Back-substituting and using the original limits is equally valid, and often safer when the substitution is awkward to invert.

Topic-Specific Study Method

Name the Inside, Then Follow It

Exam disguise	What you see on the page	What your brain must translate	Required execution
Power of a trig function	$\sin^{4}x\cos x$	$u = \sin x$ , $du = \cos x\,dx$	absorb the cosine
Exponent with a square	$x\,e^{x^{2}}$	$du = 2x\,dx$ , so a $\frac{1}{2}$ appears	adjust the constant only
Fraction with a quadratic below	$\frac{2x}{x^{2}+1}$	$u$ is the denominator	gives a logarithm
Root of a quadratic	$x\sqrt{x^{2}+4}$	$u = x^{2}+4$	power rule on $u^{1/2}$
Definite integral with a composite	limits plus an inner function	the limits must change too	convert before integrating

① Write $u$ and $du$ on their own lines. The two lines are the method, and skipping them is why students lose track of which variable is live.

② Convert the limits at the same moment you convert the integrand. A definite integral in $u$ with $x$ -limits is the single most common structural error in the topic.

Training & Error Diagnosis

Core Archetype (Simplest Form): Find $\int 2x\left(x^{2}+3\right)^{5}dx$ .

Standard Execution:

Choose. $u = x^{2}+3$ , so $du = 2x\,dx$ .

Rewrite. The integral becomes $\int u^{5}\,du$ .

Integrate. $\frac{u^{6}}{6} + C$ .

Return. $\frac{\left(x^{2}+3\right)^{6}}{6} + C$ .

Check. Differentiating gives $\left(x^{2}+3\right)^{5}\cdot 2x$ , which matches.

Exam Variation (Context / Parameter Shift): Evaluate $\int_1^{2} \frac{3x^{2}}{x^{3}+1}\,dx$ .

Solution: Let $u = x^{3}+1$ , so $du = 3x^{2}dx$ and the integrand is exactly $\frac{du}{u}$ . The limits convert: $x = 1$ gives $u = 2$ , and $x = 2$ gives $u = 9$ . So the integral is $\int_2^{9}\frac{du}{u} = \left[\ln u\right]_2^{9} = \ln 9 - \ln 2 = \ln\frac{9}{2}$ . Note the two decisions that carry the points: the limits moved with the substitution, and the absolute value in the logarithm can be dropped because both limits are positive.

Common Traps & Failure Signals:

Limits left in $x$ (Signal: an $x$ -value substituted into a $u$ -expression): convert the limits or back-substitute, never mix.
Missing $du$ factor (Signal: an $x$ left over after substituting): the derivative of the inside must be fully absorbed.
Invented factor (Signal: multiplying by $x$ to make the match work): only constants may be adjusted.

6D Cognitive Error Diagnosis (Where Your Thinking Broke):

Inside Unnamed: Write $u$ before rewriting.
Variable Mixing: You converted the integrand and kept the old limits, which is a bookkeeping failure rather than a calculus one, and it produces a wrong number with no visible symptom because every individual step looks correct.
Factor Invented: You supplied the missing $x$ to make the substitution fit, which works only when the missing factor is a constant, and the resulting answer differentiates back to something that is not the original integrand at all.

Past 10-Year AP Exam Appearance

Testing Frequency: MCQ Only; No released free-response part since 2017 requires u-substitution as a scored step; every appearance is in the multiple-choice section or inside a differential-equation solution
10-Year Total Count: 0 of 9 Exams · 0 FRQ Parts; The nine free-response sets contain no part whose rubric mentions substitution
FRQ Years Appeared: None in Released FRQs ---

Practice Examples

MCQ 1

The Straightforward Case

$\int 2x\left(x^{2}+5\right)^{3}dx =$

MCQ 2

Changing the Limits

$\int_0^{1} 3x^{2}\left(x^{3}+1\right)^{2}dx =$

MCQ 3

Recognising a Logarithm

$\int \frac{4x}{x^{2}+9}\,dx =$

FRQ 4

Substitution in Both Directions

(a) Find $\int x^{2}\sqrt{x^{3}+1}\,dx$ .

(b) Evaluate $\int_0^{2} \frac{x}{x^{2}+1}\,dx$ .

Cheatsheet

Core Reference

	Action	Failure if skipped
1	name $u$ as the inner function	nothing collapses
2	compute $du$ and solve for the leftover	an $x$ survives
3	adjust constants only	the answer differentiates back wrong
4	convert the limits (definite)	$x$ -limits on a $u$ -integrand
5	back-substitute (indefinite)	the answer is in the wrong variable

Sentences That Score

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Unit 6Integration by Substitution

Concept 4.87Built for May 2027 AP Exam

Integration by Substitution

Reviewed by ACEpath AP Math Editorial Team

Last Updated: September 2026

30-second summary

On the AP Exam

AP Calculus AB tests this skill through the representations and reasoning described below.

What you’ll see: Seldom as a free-response part of its own. It appears inside differential-equation solutions and inside multiple-choice items, usually as $\int x\,e^{x^{2}}dx$ or $\int \frac{2x}{x^{2}+1}\,dx$ , and often as one step of a longer computation.
What you’ll do: Pick $u$ as the inner function, compute $du$ , match the remaining factor, and then either back-substitute or convert the limits.
Memory hook: Name the inside, then follow it down. Whatever sits inside the parentheses or the exponent becomes $u$ , and everything else must become $du$ .
Use it on the exam: For MCQs: Differentiate the choices rather than integrating them. A candidate answer that differentiates back to the integrand is correct, and the check bypasses the whole substitution. For FRQs: Write $u$ and $du$ on separate lines before rewriting the integral. When the integral is definite, write the new limits immediately, in the same line, so they cannot be forgotten.

Concept Review

The Chain Rule, Reversed

Figure 1 — Substitution changes the variable and the interval together. The region on the $x$ -axis maps to a region on the $u$ -axis, and the limits must travel with it.

The chain rule says $\frac{d}{dx}F\left(g(x)\right) = F'\left(g(x)\right)g'(x)$ . Integrating both sides gives the substitution rule:

$\int f\left(g(x)\right)g'(x)\,dx = F\left(g(x)\right) + C, \qquad \text{with } u = g(x),\ du = g'(x)\,dx$

In practice you never write $F$ . You write $u$ , you write $du$ , and the integral becomes $\int f(u)\,du$ .

Substitution changes the variable and the interval together. The region on the $x$-axis maps to a region on the $u$-axis, and the limits must travel with it.

Choosing $u$

Integrand	Natural choice of $u$	Why
$x\,e^{x^{2}}$	$u = x^{2}$	$du = 2x\,dx$ , and the $x\,dx$ is present up to the $2$
$\frac{2x}{x^{2}+1}$	$u = x^{2}+1$	$du = 2x\,dx$ matches exactly
$\cos(3x)\,dx$	$u = 3x$	$du = 3\,dx$
$\sin^{4}x\cos x\,dx$	$u = \sin x$	$du = \cos x\,dx$
$x\sqrt{x^{2}+4}\,dx$	$u = x^{2}+4$	$du = 2x\,dx$ , present up to a factor

The pattern in every row is the same: the derivative of the inside appears somewhere in the integrand, possibly multiplied by a constant you can move.

Adjusting a Constant Is Legal; Adjusting a Variable Is Not

$\int x\,e^{x^{2}}dx = \frac{1}{2}\int e^{u}\,du = \frac{1}{2}e^{u} + C = \frac{1}{2}e^{x^{2}} + C$

Definite Integrals: Change the Limits

$\int_0^{1} x\,e^{x^{2}}dx = \frac{1}{2}\int_{0}^{1} e^{u}\,du = \frac{1}{2}\left[e^{u}\right]_0^{1} = \frac{1}{2}(e - 1)$

When Substitution Fails

Common Misconception Check

Myth 1: “Substitution works on any composite integrand.” Truth: It works when the inner derivative is present. Without that factor the method has nothing to absorb.
Myth 2: “After substituting, the old limits still apply.” Truth: The limits are $x$ -values and the integrand is now a $u$ -expression. Both must speak the same variable.
Myth 3: “You must always convert the limits.” Truth: Back-substituting and using the original limits is equally valid, and often safer when the substitution is awkward to invert.

Topic-Specific Study Method

Name the Inside, Then Follow It

Exam disguise	What you see on the page	What your brain must translate	Required execution
Power of a trig function	$\sin^{4}x\cos x$	$u = \sin x$ , $du = \cos x\,dx$	absorb the cosine
Exponent with a square	$x\,e^{x^{2}}$	$du = 2x\,dx$ , so a $\frac{1}{2}$ appears	adjust the constant only
Fraction with a quadratic below	$\frac{2x}{x^{2}+1}$	$u$ is the denominator	gives a logarithm
Root of a quadratic	$x\sqrt{x^{2}+4}$	$u = x^{2}+4$	power rule on $u^{1/2}$
Definite integral with a composite	limits plus an inner function	the limits must change too	convert before integrating

① Write $u$ and $du$ on their own lines. The two lines are the method, and skipping them is why students lose track of which variable is live.

② Convert the limits at the same moment you convert the integrand. A definite integral in $u$ with $x$ -limits is the single most common structural error in the topic.

Training & Error Diagnosis

Core Archetype (Simplest Form): Find $\int 2x\left(x^{2}+3\right)^{5}dx$ .

Standard Execution:

Choose. $u = x^{2}+3$ , so $du = 2x\,dx$ .

Rewrite. The integral becomes $\int u^{5}\,du$ .

Integrate. $\frac{u^{6}}{6} + C$ .

Return. $\frac{\left(x^{2}+3\right)^{6}}{6} + C$ .

Check. Differentiating gives $\left(x^{2}+3\right)^{5}\cdot 2x$ , which matches.

Exam Variation (Context / Parameter Shift): Evaluate $\int_1^{2} \frac{3x^{2}}{x^{3}+1}\,dx$ .

Solution: Let $u = x^{3}+1$ , so $du = 3x^{2}dx$ and the integrand is exactly $\frac{du}{u}$ . The limits convert: $x = 1$ gives $u = 2$ , and $x = 2$ gives $u = 9$ . So the integral is $\int_2^{9}\frac{du}{u} = \left[\ln u\right]_2^{9} = \ln 9 - \ln 2 = \ln\frac{9}{2}$ . Note the two decisions that carry the points: the limits moved with the substitution, and the absolute value in the logarithm can be dropped because both limits are positive.

Common Traps & Failure Signals:

Limits left in $x$ (Signal: an $x$ -value substituted into a $u$ -expression): convert the limits or back-substitute, never mix.
Missing $du$ factor (Signal: an $x$ left over after substituting): the derivative of the inside must be fully absorbed.
Invented factor (Signal: multiplying by $x$ to make the match work): only constants may be adjusted.

6D Cognitive Error Diagnosis (Where Your Thinking Broke):

Inside Unnamed: Write $u$ before rewriting.
Variable Mixing: You converted the integrand and kept the old limits, which is a bookkeeping failure rather than a calculus one, and it produces a wrong number with no visible symptom because every individual step looks correct.
Factor Invented: You supplied the missing $x$ to make the substitution fit, which works only when the missing factor is a constant, and the resulting answer differentiates back to something that is not the original integrand at all.

Past 10-Year AP Exam Appearance

Testing Frequency: MCQ Only; No released free-response part since 2017 requires u-substitution as a scored step; every appearance is in the multiple-choice section or inside a differential-equation solution
10-Year Total Count: 0 of 9 Exams · 0 FRQ Parts; The nine free-response sets contain no part whose rubric mentions substitution
FRQ Years Appeared: None in Released FRQs ---

Practice Examples

MCQ 1

The Straightforward Case

$\int 2x\left(x^{2}+5\right)^{3}dx =$

MCQ 2

Changing the Limits

$\int_0^{1} 3x^{2}\left(x^{3}+1\right)^{2}dx =$

MCQ 3

Recognising a Logarithm

$\int \frac{4x}{x^{2}+9}\,dx =$

FRQ 4

Substitution in Both Directions

(a) Find $\int x^{2}\sqrt{x^{3}+1}\,dx$ .

(b) Evaluate $\int_0^{2} \frac{x}{x^{2}+1}\,dx$ .

Cheatsheet

Core Reference

	Action	Failure if skipped
1	name $u$ as the inner function	nothing collapses
2	compute $du$ and solve for the leftover	an $x$ survives
3	adjust constants only	the answer differentiates back wrong
4	convert the limits (definite)	$x$ -limits on a $u$ -integrand
5	back-substitute (indefinite)	the answer is in the wrong variable

Sentences That Score

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