Do I always need the absolute value in |u|?

Only when u can be negative. (x2 + 1) needs none, because x2 + 1 is always positive, but 1x\,dx = |x| + C does.

Unit 6 · Topic 6.14 · about 30 minutes · 7 parts · 12 scored checks, 3 of them multiple choice

AP Calculus AB 6.14: Selecting Techniques for Antidifferentiation

Q: Can two different-looking answers both be right?

Yes, when they differ by a constant. 12(x + 1)2 + C and 12x2 + x + C are both antiderivatives of x + 1; they differ by 12, which the + C absorbs.

Q: Why does the arctangent keep showing up?

Because ddx x = 11 + x2. Any constant over a square plus a positive constant can be brought to that form, and completing the square and factoring out the constant are the moves that do it.

Built by Tanner Crow from College Board's course framework · Every answer key independently checked

Unit 6: 15 to 20% of the multiple-choice section

Key vocabulary

Term	What it means	The exam trap
Antiderivative	A function $F$ whose derivative is $f$ , so that $F^{'} (x) = f (x)$ .	Differentiate your answer before moving on; if it does not give back the integrand exactly, the technique or the algebra slipped.
Indefinite integral	$\int f (x) d x = F (x) + C$ , where $F^{'} (x) = f (x)$ and $C$ is any constant.	Write $+ C$ on every indefinite integral, including one found by substitution or by division.
Substitution of variables	A technique for finding antiderivatives that replaces an inside function $g (x)$ with $u$ and $g^{'} (x) d x$ with $d u$ , so the integral is written in $u$ alone.	On a definite integral, change the limits to values of $u$ , or return to $x$ before you evaluate.
Long division	Dividing the bottom polynomial into the top to write a fraction as a polynomial plus a fraction whose top has lower degree.	Check the division by multiplying back: the quotient times the divisor, plus the remainder, must equal the original top.
Completing the square	Rewriting a quadratic $x^{2} + b x + c$ as a square plus a constant, ${(x + \frac{b}{2})}^{2} + c - \frac{b^{2}}{4}$ .	Add and subtract the same number so the quadratic keeps its value, and expand back to check.

Graph of y = 1/(x^2 - 4x + 5) beside a dashed graph of y = 1/(1 + x^2). Completing the square writes the first bottom as (x - 2)^2 + 1, so the first curve is the second moved right 2, with its peak of 1 at x = 2. The region under the first curve from x = 1 to x = 3 is shaded and has area 1.571.

Part 1 Think first Which tool opens this integral?

Commit to a guess before you read on. Guessing first, even wrongly, makes the lesson stick better.

Predict first

Meltwater runs into a rain barrel at $r (t) = \frac{6 t}{t^{2} + 9}$ liters per hour, $t$ hours after sunrise. To find how much water collects in the first 4 hours you need an antiderivative of $r$ . Which approach finds one?

In short. Selecting a technique for antidifferentiation means reading the form of an integrand and choosing the move that turns it into one you can antidifferentiate: a rewrite and a basic rule, a substitution, long division, or completing the square.

Look for a basic rule first, after writing roots as powers, multiplying out products and splitting a fraction over a single term.
A substitution $u = g (x)$ works when $g^{'} (x)$ is also a factor of the integrand, up to a constant.
A fraction with no such factor may need long division (the top's degree at least the bottom's) or completing the square (a quadratic bottom under a constant).

Part 2 Learn it Read the integrand, then choose

6a. Rewrite until a basic rule fits

Every antiderivative you know is a derivative you know, read backward: $\frac{d}{d x} (\frac{x^{n + 1}}{n + 1}) = x^{n}$ for $n \neq - 1$ , $\frac{d}{d x} \ln | x | = \frac{1}{x}$ , $\frac{d}{d x} \sin x = \cos x$ , $\frac{d}{d x} e^{x} = e^{x}$ , $\frac{d}{d x} \arctan x = \frac{1}{1 + x^{2}}$ , and the rest of your derivative rules. So the first question for any integrand is whether it is already a sum of those forms, or can be made into one with algebra.

Three rewrites do most of that work. Write roots and reciprocals as powers: $\sqrt[3]{x} = x^{1 / 3}$ and $\frac{4}{x^{2}} = 4 x^{- 2}$ . Multiply out a product: $(x + 1) (x - 3) = x^{2} - 2 x - 3$ . Split a fraction whose bottom is a single term: $\frac{x^{2} - 5}{x} = x - \frac{5}{x}$ . Then integrate term by term:

\int \frac{x^{2} - 5}{x} d x = \int (x - \frac{5}{x}) d x = \frac{x^{2}}{2} - 5 \ln | x | + C .

There is no quotient rule for integrals and no product rule either. The antiderivative of a fraction is not the antiderivative of its top over the antiderivative of its bottom. Rewrite first, and check any answer by differentiating it: its derivative must be the integrand, exactly.

Quick check · Multiple choice

No calculator. Which of these is $\int \frac{x + 2}{\sqrt{x}} d x$ ?

6b. Substitution: an inside function and its derivative

Substitution runs the chain rule backward. If the integrand is $f (g (x)) g^{'} (x)$ , then with $u = g (x)$ and $d u = g^{'} (x) d x$ it becomes $\int f (u) d u$ .

So look for an inside function whose derivative also sits in the integrand as a factor, up to a constant. In $\int x^{2} \cos (x^{3}) d x$ the inside function is $x^{3}$ . Its derivative, $3 x^{2}$ , differs from the $x^{2}$ outside only by the factor 3, so with $u = x^{3}$ , $x^{2} d x = \frac{1}{3} d u$ and

\int x^{2} \cos (x^{3}) d x = \frac{1}{3} \int \cos u d u = \frac{1}{3} \sin (x^{3}) + C .

Test a choice of $u$ by asking whether every $x$ leaves, including the one in $d x$ . A constant factor can be fixed by multiplying and dividing. A leftover $x$ has to be written in terms of $u$ too. When that is easy, as $x = u - 1$ is for $u = x + 1$ , carry on; when it is not, try a different $u$ or a different technique.

For a definite integral, the limits change with the variable. If $u = x^{3}$ and $x$ runs from 0 to 2, then $u$ runs from $0^{3} = 0$ to $2^{3} = 8$ , and you never need to return to $x$ . Keeping the $x$ limits on a $d u$ integral gives a wrong number.

Quick check · Calculate

No calculator. Evaluate $\int_{0}^{1} 6 x (x^{2} + 1)^{2} d x$ .

Your answer

6c. Rearrange first: long division and completing the square

Some fractions fit no basic rule and offer no inside function whose derivative is a factor. Two rearrangements turn them into forms that do.

Long division, when the top's degree is at least the bottom's. Divide to get a polynomial plus a fraction whose top has lower degree:

\frac{x^{2} + 2}{x^{2} + 1} = 1 + \frac{1}{x^{2} + 1}, so \int \frac{x^{2} + 2}{x^{2} + 1} d x = x + \arctan x + C .

Completing the square, when a constant sits over a quadratic with no real zeros. Write the quadratic as a square plus a positive constant, and the arctangent rule fits after a shift:

x^{2} + 10 x + 26 = (x + 5)^{2} + 1, so \int \frac{1}{x^{2} + 10 x + 26} d x = \arctan (x + 5) + C .

Before either one, look for the shortcut. If the top is a constant times the derivative of the bottom, as in $\frac{2 x + 10}{x^{2} + 10 x + 26}$ , the substitution $u = x^{2} + 10 x + 26$ finishes it, and the answer is a logarithm, $\ln (x^{2} + 10 x + 26) + C$ . A constant on top can never be the derivative of a quadratic bottom, which always has an $x$ in it.

Quick check · Sort it

No calculator. Sort each integrand by what it needs before an antiderivative can be written. Tap a card, then tap its bin.

Long division first

Completing the square first

Substitution, no rearranging

Common mistakes

The mistake	Why it is wrong	Do this instead
Antidifferentiating a fraction by antidifferentiating its top and its bottom separately.	There is no quotient rule for integrals. The derivative of $\frac{F (x)}{G (x)}$ is not $\frac{F^{'} (x)}{G^{'} (x)}$ , so the reverse fails too.	Rewrite first: split over a single-term bottom, divide, complete the square or substitute. Then differentiate your answer to check it.
Writing a logarithm for every fraction, as in $\int \frac{1}{x^{2} + 9} d x = \ln (x^{2} + 9) + C$ .	The derivative of $\ln (x^{2} + 9)$ is $\frac{2 x}{x^{2} + 9}$ , not $\frac{1}{x^{2} + 9}$ . A logarithm fits only when the top is the bottom's derivative, up to a constant.	Ask whether the top is a constant times the bottom's derivative. If the top is a constant and the bottom a quadratic, look toward the arctangent.
Keeping the $x$ limits after a substitution in a definite integral.	The new integral is in $u$ , and $u$ runs from $g (a)$ to $g (b)$ , not from $a$ to $b$ .	Change both limits when you write $d u$ , or return to $x$ before you evaluate.

Part 3 Watch it done Choosing the move, one step at a time

Open the steps one at a time. The last move is yours.

Find $\int \frac{1}{x^{2} + 4 x + 8} d x$ .

Predict first

Before the steps open: which move should come first for $\int \frac{1}{x^{2} + 4 x + 8} d x$ ?

Step 1 · Read the integrand
The top is the constant 1. The bottom's derivative is $2 x + 4$ , which is not a constant multiple of 1, so $u = x^{2} + 4 x + 8$ would leave an $x$ behind. The top's degree is already lower than the bottom's, so there is nothing to divide. The power rule does not apply either, because the base $x^{2} + 4 x + 8$ is not $x$ . A constant over a quadratic: complete the square.
Step 2 · Complete the square
$x^{2} + 4 x + 8 = (x^{2} + 4 x + 4) + 4 = (x + 2)^{2} + 4$ , so the integral is $\int \frac{1}{(x + 2)^{2} + 4} d x$ .
Step 3 · Make a 1 appear beside the square
$(x + 2)^{2} + 4 = 4 ({(\frac{x + 2}{2})}^{2} + 1)$ , so the integral is $\frac{1}{4} \int \frac{1}{1 + {(\frac{x + 2}{2})}^{2}} d x$ .
Why this step?
Why factor out the 4, instead of writing $\arctan (x + 2) + C$ straight after completing the square?
The rule is $\int \frac{1}{1 + u^{2}} d u = \arctan u + C$ , with 1, not 4, beside the square. The derivative of $\arctan (x + 2)$ is $\frac{1}{1 + (x + 2)^{2}}$ , a different function from $\frac{1}{4 + (x + 2)^{2}}$ . Factoring out the 4 puts the integrand in the rule's form.
Your step · Fill the statement
No calculator. Finish it. Substitute for the quantity being squared, keeping track of what $d x$ becomes. Tap a blank, then tap an answer.

$\int \frac{1}{x^{2} + 4 x + 8} d x$ is times the arctangent of , plus $C$ .

Part 4 Try it Two integrals you have not seen

Try it · Decide and justify

No calculator. A classmate says $\int \frac{e^{x}}{1 + e^{2 x}} d x$ can be found with one substitution that turns it into the arctangent rule. Is the classmate right?

Which reason supports your answer?

Try it · Calculate

No calculator. Evaluate $\int_{0}^{π / 2} \cos^{3} x \sin x d x$ . Enter your answer as a decimal.

Your answer

Part 5 Explain it Write it out, then score it

Free response is written by hand on the exam, so do this one on paper. Nothing typed here leaves the page. When you finish, open the model answer and score yourself point by point.

Free response is 50% of the exam score. This one is shaped like Question 3, from free-response Part B (calculator not permitted).

Free-response practice · Question 3 style

No calculator. Water is pumped into a tank at the rate $R (t) = \frac{t^{2} + 3 t + 4}{t + 1}$ liters per minute, where $t$ is measured in minutes and $0 \leq t \leq 3$ .

(a) Write $R (t)$ as a sum of terms, each of which can be antidifferentiated with a basic rule. Show the work that leads to your answer.

(b) Find the total amount of water, in liters, pumped into the tank from $t = 0$ to $t = 3$ . Show the work that leads to your answer.

(c) A student begins part (b) with the substitution $u = t^{2} + 3 t + 4$ . Explain why this substitution does not simplify the integral.

Your answer stays on this page. It is never sent or saved.

Model answer

(a) Dividing $t + 1$ into $t^{2} + 3 t + 4$ gives $t^{2} + 3 t + 4 = (t + 1) (t + 2) + 2$ , so $R (t) = t + 2 + \frac{2}{t + 1}$ .

(b) $\int_{0}^{3} R (t) d t = {[\frac{t^{2}}{2} + 2 t + 2 \ln (t + 1)]}_{0}^{3} = (\frac{9}{2} + 6 + 2 \ln 4) - (0 + 0 + 2 \ln 1) = \frac{21}{2} + 2 \ln 4$ liters.

(c) With $u = t^{2} + 3 t + 4$ , $d u = (2 t + 3) d t$ , and $2 t + 3$ is not a factor of $R (t)$ , even up to a constant. So $R (t) d t = \frac{u}{(t + 1) (2 t + 3)} d u$ still contains $t$ , and removing it would mean solving the quadratic for $t$ , which gives a harder integral than the one the rewriting in part (a) leads to in part (b).

Score your own answer. This is a self-check, not a grade. Tick each point your answer makes.

Divides to get $R (t) = t + 2 + \frac{2}{t + 1}$ , showing the division or $(t + 1) (t + 2) + 2$ .
Writes the antiderivative $\frac{t^{2}}{2} + 2 t + 2 \ln (t + 1)$ .
Uses the limits 0 and 3 and gets $\frac{21}{2} + 2 \ln 4$ liters, with units.
Explains that $2 t + 3$ is not a factor of $R (t)$ , even up to a constant, so $t$ is left after substituting, and removing it would mean solving the quadratic for $t$ , a harder integral.

Part 6 Check yourself A function known only by its table

The two case questions use the table. Read every value you need from it.

The function $f$ is differentiable, and its derivative $f^{'}$ is continuous. Selected values of $f$ and $f^{'}$ are in the table.

Selected values of f and its derivative
$x$	$f (x)$	$f^{'} (x)$
1	2	1
2	3	1
3	4	3
4	7	4
6	10	1

Case question 1 · Calculate

No calculator. Evaluate $\int_{1}^{3} f (x) f^{'} (x) d x$ .

Your answer

Case question 2 · Multiple choice

No calculator. What is the value of $\int_{1}^{3} f^{'} (2 x) d x$ ?

The rest stand on their own and do not use the case.

Check yourself · Spot the error

No calculator. A student finds $\int \frac{1}{x^{2} - 8 x + 17} d x$ . Tap the first line with a mistake.

Check yourself · Match

No calculator. These four integrands look alike, but they do not all take the same approach. Tap an integrand, then its antiderivative. Differentiating a candidate is the check.

Part 7 Keep it Earlier ideas this lesson stands on

Substitution is the chain rule run backward, and every definite integral here rests on the Fundamental Theorem. Both come back now.

Bring it back · Calculate

No calculator. Let $g (x) = \int_{2}^{x} \sqrt{t^{3} + 1} d t$ . What is $g^{'} (2)$ ?

Your answer

Bring it back · Multiple choice

No calculator. What is $\frac{d}{d x} \sin (x^{3})$ ?

In three lines

Rewrite first, then use a basic rule: roots as powers, products multiplied out, a fraction split over a single-term bottom.
Substitute when an inside function's derivative is also a factor, up to a constant, and change the limits of a definite integral with it.
Divide when the top's degree is at least the bottom's, and complete the square when a constant sits over a quadratic.

Unit 6 check: 11 questions from every lesson in Unit 6

Exam strategy for this topic

Section I Part A, Multiple Choice (calculator not permitted): 29 questions in 62 minutes.
Section I Part B, Multiple Choice (graphing calculator required): 13 questions in 38 minutes.
Section II Part A, Free Response (graphing calculator required): 2 questions in 30 minutes.
Section II Part B, Free Response (calculator not permitted): 4 questions in 60 minutes.

On a no-calculator integral, write the substitution you choose ( $u = \dots$ and $d u = \dots d x$ ) and the new limits before you integrate, and put $+ C$ on every indefinite integral. Before you commit to a technique, differentiate your candidate antiderivative: it has to give back the integrand.

Questions students ask

How do I know which technique to try first?

Try them in order of effort. Can algebra make the integrand a sum of basic forms? Is there an inside function whose derivative is also a factor? If it is a fraction with neither, is the top's degree at least the bottom's (divide), or is it a constant over a quadratic (complete the square)?

Do I always need the absolute value in

\ln | u |

Only when $u$ can be negative. $\ln (x^{2} + 1)$ needs none, because $x^{2} + 1$ is always positive, but $\int \frac{1}{x} d x = \ln | x | + C$ does.

Can two different-looking answers both be right?

Yes, when they differ by a constant. $\frac{1}{2} (x + 1)^{2} + C$ and $\frac{1}{2} x^{2} + x + C$ are both antiderivatives of $x + 1$ ; they differ by $\frac{1}{2}$ , which the $+ C$ absorbs.

What if no technique works?

Many functions have no antiderivative that can be written with the usual functions, $e^{- x^{2}}$ among them. A definite integral of such a function is found with a graphing calculator or estimated with a Riemann sum.

Why does the arctangent keep showing up?

Because $\frac{d}{d x} \arctan x = \frac{1}{1 + x^{2}}$ . Any constant over a square plus a positive constant can be brought to that form, and completing the square and factoring out the constant are the moves that do it.

Course alignment, for teachers

AP Calculus AB topic 6.14, Unit 6: Integration and Accumulation of Change.

Skill 1.C: Identify an appropriate mathematical rule or procedure based on the classification of a given expression (e.g., Use the chain rule to find the derivative of a composite function).

This lesson maps the topic's 0 learning objectives, since the course framework gives it no essential knowledge statements, to the part that teaches each one: 0 are checked by a scored question. The map is in the lesson file.