Unit 6 · Topic 6.10 · about 30 minutes · 7 parts · 12 scored checks, 3 of them multiple choice

AP Calculus AB 6.10: Integrating Functions Using Long Division and Completing the Square

Q: Do I need the absolute value in |x + 2|?

In an indefinite integral, yes: |u| is an antiderivative of 1u for negative u as well as positive. Over an interval where x + 2 stays positive, such as 0 x 3, (x + 2) gives the same values.

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

What you'll be able to do

Find indefinite integrals and evaluate definite integrals of fractions by first rewriting them with long division or by completing the square.

Key vocabulary

Term	What it means	The exam trap
Antiderivative	A function $F$ whose derivative is $f$ , so that $F^{'} (x) = f (x)$ .	Differentiate a proposed antiderivative before you trust it; if the result is not exactly the integrand, the antiderivative is wrong.
Indefinite integral	$\int f (x) d x = F (x) + C$ , where $F^{'} (x) = f (x)$ and $C$ is any constant.	Every indefinite integral ends in $+ C$ . Without it you have written one antiderivative, not all of them.
Equivalent forms	Two expressions that are equal for every input where both are defined, such as $\frac{x^{2} + 3 x + 5}{x + 1}$ and $x + 2 + \frac{3}{x + 1}$ .	A rewrite is allowed only if it equals the original. Recombine or multiply back to check it before you integrate.
Long division	Dividing one polynomial by another to write a fraction as a quotient plus a remainder over the divisor.	Subtract the whole product at each step, term by term; one sign slip changes the remainder and everything that comes from it.
Completing the square	Rewriting $x^{2} + b x + c$ as ${(x + \frac{b}{2})}^{2} + c - \frac{b^{2}}{4}$ .	Whatever you add inside the square, take back outside it, so the expression keeps its value.

Graph of y = (x squared + 3x + 5)/(x + 1) for x from 0 to 4, lying above the dashed line y = x + 2. The shaded gap between them from x = 0 to x = 2 is the remainder piece 3/(x + 1), with area 3.296.

Part 1 Think first How much water in two minutes?

Guess first. Committing to an answer, even a wrong one, makes the explanation stick.

Predict first

A pump fills a tank at a rate of $r (t) = \frac{t^{2} + 2 t + 3}{t + 1}$ liters per minute, $t$ minutes after it starts. At the start the rate is 3 liters per minute, and at $t = 2$ it is about 3.7. About how much water goes into the tank in the first 2 minutes?

In short. Integrating with long division or completing the square means rewriting a fraction as an equal sum of pieces that each have a known antiderivative, then integrating piece by piece.

When the top's degree is at least the bottom's, divide: the fraction equals a quotient plus a remainder over the divisor.
When a constant sits over a quadratic with no real zeros, complete the square and look for an inverse tangent.
A definite integral uses the same rewrite, then $F (b) - F (a)$ with the antiderivative.

Part 2 Learn it Rewrite first, then integrate

6a. Top-heavy fractions: divide first

There is no quotient rule for integrals. When the top of a fraction has degree at least the degree of the bottom, divide first. Long division rewrites the fraction as a polynomial plus a remainder over the divisor:

\frac{top}{bottom} = quotient + \frac{remainder}{bottom} .

Take $\frac{x^{2} + 3 x + 5}{x + 1}$ . $x + 1$ goes into $x^{2}$ exactly $x$ times. Subtract $x (x + 1) = x^{2} + x$ and $2 x + 5$ is left. $x + 1$ goes into $2 x$ twice. Subtract $2 (x + 1) = 2 x + 2$ and 3 is left. So

\frac{x^{2} + 3 x + 5}{x + 1} = x + 2 + \frac{3}{x + 1} .

Check it by multiplying back: $(x + 1) (x + 2) + 3 = x^{2} + 3 x + 5$ . The two forms are equivalent, so they have the same antiderivatives, and now every piece has a rule:

\int \frac{x^{2} + 3 x + 5}{x + 1} d x = \frac{x^{2}}{2} + 2 x + 3 \ln | x + 1 | + C .

The graph at the top shows the rewrite. The curve is the line $y = x + 2$ plus a gap, $\frac{3}{x + 1}$ , that shrinks as $x$ grows. The remainder piece is the one that turns into a logarithm.

Quick check · Multiple choice

No calculator. Which of the following is $\int \frac{x^{2} + 5 x + 2}{x + 2} d x$ ?

6b. A quadratic with no zeros: complete the square

Some fractions have a constant on top and a quadratic with no real zeros on the bottom, such as $\frac{1}{x^{2} + 8 x + 17}$ . Division does nothing here, because the top already has the lower degree. Complete the square instead: take half the coefficient of $x$ , square it, and split the constant around it.

x^{2} + 8 x + 17 = (x^{2} + 8 x + 16) + 1 = (x + 4)^{2} + 1.

Now the integrand has the shape $\frac{1}{u^{2} + 1}$ with $u = x + 4$ and $d u = d x$ , and you know $\frac{d}{d u} \arctan u = \frac{1}{1 + u^{2}}$ . So

\int \frac{d x}{x^{2} + 8 x + 17} = \arctan (x + 4) + C .

When the constant left over is not 1, the inverse tangent picks up a factor. For $a > 0$ , $\frac{d}{d u} \arctan (\frac{u}{a}) = \frac{a}{a^{2} + u^{2}}$ , which is $a$ times too big, so

\int \frac{d u}{u^{2} + a^{2}} = \frac{1}{a} \arctan (\frac{u}{a}) + C .

For example, $x^{2} - 2 x + 10 = (x - 1)^{2} + 9$ , so $a = 3$ and

\int \frac{d x}{x^{2} - 2 x + 10} = \frac{1}{3} \arctan (\frac{x - 1}{3}) + C .

Differentiate an answer like this to check it.

Quick check · Match

No calculator. Complete the square in each denominator. Tap an integral, then its antiderivative.

6c. Choosing the rewrite, and definite integrals

Look at a fraction before you start on it. Three questions sort most of them.

Is the top's degree at least the bottom's? Divide first.
Is the top a constant multiple of the bottom's derivative? Substitute $u$ for the bottom, as it stands.
Is the top a constant and the bottom a quadratic with no real zeros? Complete the square.

A definite integral needs nothing new: rewrite, find an antiderivative $F$ , then use $\int_{a}^{b} f (x) d x = F (b) - F (a)$ . For example, $\frac{x^{2} + 2}{x^{2} + 1}$ has top and bottom of the same degree, so divide: $\frac{x^{2} + 2}{x^{2} + 1} = 1 + \frac{1}{x^{2} + 1}$ . Then

\int_{0}^{1} \frac{x^{2} + 2}{x^{2} + 1} d x = (1 + \arctan 1) - (0 + \arctan 0) = 1 + \frac{π}{4} .

Write the antiderivative, $x + \arctan x$ , before any numbers go in, and put parentheses around each value, so the minus sign applies to the whole of $F (a)$ . Leave the answer exact, as $1 + \frac{π}{4}$ , unless a decimal is asked for.

Quick check · Sort it

No calculator. Sort each integrand by the method that handles it most directly, as it is written. Tap a card, then a bin.

Divide first

Substitute for the bottom

Complete the square

Common mistakes

The mistake	Why it is wrong	Do this instead
Integrating the top and the bottom of a fraction separately.	The integral of a quotient is not the quotient of the integrals. Differentiate the result and you do not get the integrand back.	Rewrite the fraction as a sum first, by long division or by completing the square, then integrate each piece.
Dropping the remainder after long division, or giving it the wrong sign.	The fraction equals the quotient plus the remainder over the divisor. The remainder piece is part of the integrand, and over a linear divisor it becomes a logarithm.	Multiply back before integrating: divisor times quotient, plus remainder, must give the top exactly.
Writing $\arctan (\frac{u}{a}) + C$ for $\int \frac{d u}{u^{2} + a^{2}}$ , with no factor in front.	The derivative of $\arctan (\frac{u}{a})$ is $\frac{a}{u^{2} + a^{2}}$ , which is $a$ times the integrand.	Put $\frac{1}{a}$ in front, and differentiate your answer to check it.

Part 3 Watch it done A definite integral by completing the square

Open one step at a time. The last line is yours.

Evaluate $\int_{2}^{4} \frac{d x}{x^{2} - 4 x + 8}$ without a calculator.

Step 1 · Complete the square
Half of $- 4$ is $- 2$ , and $(- 2)^{2} = 4$ , so $x^{2} - 4 x + 8 = (x^{2} - 4 x + 4) + 4 = (x - 2)^{2} + 4$ .
Step 2 · Find an antiderivative
With $u = x - 2$ , $d u = d x$ and $a = 2$ : $\int \frac{d x}{(x - 2)^{2} + 4} = \frac{1}{2} \arctan (\frac{x - 2}{2}) + C$ .
Why this step?
Why is there a $\frac{1}{2}$ in front of the inverse tangent?
Differentiate without it: $\frac{d}{d x} \arctan (\frac{x - 2}{2}) = \frac{1}{1 + {(\frac{x - 2}{2})}^{2}} \cdot \frac{1}{2} = \frac{2}{(x - 2)^{2} + 4}$ , which is twice the integrand. The $\frac{1}{2}$ cancels that 2.
Step 3 · Apply the Fundamental Theorem
$\int_{2}^{4} \frac{d x}{x^{2} - 4 x + 8} = \frac{1}{2} \arctan (\frac{4 - 2}{2}) - \frac{1}{2} \arctan (\frac{2 - 2}{2}) = \frac{1}{2} \arctan 1 - \frac{1}{2} \arctan 0$ .
Your step · Calculate

No calculator. Finish it. $\int_{2}^{4} \frac{d x}{x^{2} - 4 x + 8} = \frac{1}{2} \arctan 1 - \frac{1}{2} \arctan 0$ , and that value is $k π$ . What is $k$ ? Give an exact decimal.

Your answer

Part 4 Try it Two new integrands

Try it · Calculate

No calculator. The exact value of $\int_{0}^{3} \frac{x^{2} + 5 x + 7}{x + 2} d x$ can be written as $a + \ln \frac{5}{2}$ . What is $a$ ? Give an exact decimal.

Your answer

Try it · Decide and justify

No calculator. A classmate says $\int \frac{x^{2}}{x^{2} + 1} d x$ cannot be found with the rules you know, because the top and the bottom have the same degree and there is no quotient rule for integrals. Is the classmate right?

Which reason supports your answer?

Part 5 Explain it Write it out by hand

Free response on the exam is handwritten, so work this on paper. Nothing you type here leaves this page. When you have an answer, open the model and score yourself against each point.

Free response is 50% of the exam score. This one is shaped like Question 6, from free-response Part B (calculator not permitted).

Free-response practice · Question 6 style

No calculator. Water is pumped into a tank at a rate of $W (t) = \frac{t^{2} + 4 t + 5}{t + 2}$ liters per minute for $0 \leq t \leq 4$ , where $t$ is measured in minutes.

(a) Use long division to write $W (t)$ as a polynomial plus a fraction whose numerator is a constant.

(b) Find the total amount of water, in liters, pumped into the tank from $t = 0$ to $t = 2$ . Show the antiderivative you use.

(c) From $t = 2$ to $t = 4$ , water also drains out of the tank at a rate of $D (t) = \frac{8}{t^{2} - 4 t + 8}$ liters per minute. Find the total amount of water, in liters, that drains out over that interval.

(d) From $t = 2$ to $t = 4$ , does the amount of water in the tank increase or decrease? Justify your answer.

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

Model answer

(a) $t^{2} + 4 t + 5 = (t + 2) (t + 2) + 1$ , so $W (t) = t + 2 + \frac{1}{t + 2}$ .

(b) $\int_{0}^{2} W (t) d t = {[\frac{t^{2}}{2} + 2 t + \ln | t + 2 |]}_{0}^{2} = (2 + 4 + \ln 4) - (0 + 0 + \ln 2) = 6 + \ln 2$ liters.

(c) $t^{2} - 4 t + 8 = (t - 2)^{2} + 4$ , so $\int_{2}^{4} \frac{8}{(t - 2)^{2} + 4} d t = {[4 \arctan (\frac{t - 2}{2})]}_{2}^{4} = 4 \arctan 1 - 4 \arctan 0 = π$ liters.

(d) Water pumped in from $t = 2$ to $t = 4$ : $\int_{2}^{4} W (t) d t = (8 + 8 + \ln 6) - (2 + 4 + \ln 4) = 10 + \ln \frac{3}{2}$ liters, which is more than 10 liters because $\ln \frac{3}{2} > 0$ . Only $π$ liters, less than 4, drains out. More water enters than leaves, so the amount of water in the tank increases.

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

Divides to get $W (t) = t + 2 + \frac{1}{t + 2}$ .
Writes the antiderivative $\frac{t^{2}}{2} + 2 t + \ln | t + 2 |$ and gets $6 + \ln 2$ liters.
Completes the square to $(t - 2)^{2} + 4$ , uses $4 \arctan (\frac{t - 2}{2})$ , and gets $π$ liters.
Finds $10 + \ln \frac{3}{2}$ liters pumped in from $t = 2$ to $t = 4$ .
Says the amount increases, because $10 + \ln \frac{3}{2}$ liters in is more than $π$ liters out.

Part 6 Check yourself One fraction, two rewrites

Both case questions use the function and the division in the table. Answer them from it.

A student divides $x^{3} + 3 x^{2} + 4 x + 5$ by $x^{2} + 2 x + 2$ to integrate $h (x) = \frac{x^{3} + 3 x^{2} + 4 x + 5}{x^{2} + 2 x + 2}$ . The table shows her long division, one step per row.

Long division of $x^{3} + 3 x^{2} + 4 x + 5$ by $x^{2} + 2 x + 2$
Step	Next term of the quotient	Subtract	Left over
1	$x$	$x^{3} + 2 x^{2} + 2 x$	$x^{2} + 2 x + 5$
2	$1$	$x^{2} + 2 x + 2$	$3$

Case question 1 · Multiple choice

No calculator. Which of the following is $\int h (x) d x$ ?

Case question 2 · Calculate

No calculator. The exact value of $\int_{- 1}^{0} h (x) d x$ can be written as $a + \frac{3 π}{4}$ . What is $a$ ? Give an exact decimal.

Your answer

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

Check yourself · Spot the error

No calculator. A student finds $\int \frac{d x}{x^{2} + 10 x + 29}$ . Tap the first line with a mistake.

Check yourself · Put it in order

No calculator. Put the steps for evaluating $\int_{1}^{2} \frac{x^{2} + x + 1}{x + 1} d x$ in order. Tap them in sequence.

Part 7 Keep it Bring back earlier ideas

This lesson leans on substitution and on the derivative of the inverse tangent. Recalling them now keeps them ready.

Bring it back · Calculate

No calculator. Evaluate $\int_{0}^{1} 2 x (x^{2} + 1)^{3} d x$ . Give an exact decimal.

Your answer

Bring it back · Multiple choice

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

In three lines

When the top's degree is at least the bottom's, long division turns a fraction into a polynomial plus a remainder over the divisor.
A constant over a quadratic with no real zeros becomes $\frac{1}{u^{2} + a^{2}}$ by completing the square, and integrates to $\frac{1}{a} \arctan (\frac{u}{a}) + C$ .
For a definite integral, rewrite, find the antiderivative, then subtract its value at the lower limit from its value at the upper limit.

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.

Without a calculator, write the rewritten integrand and the antiderivative before any numbers, and end every indefinite integral with $+ C$ . For a definite integral, show $F (b) - F (a)$ with parentheses around each value, and leave an answer such as $6 + \ln 2$ or $\frac{π}{8}$ exact unless the question asks for a decimal.

Questions students ask

How do I know whether to divide or to complete the square?

Compare degrees first. If the top's degree is at least the bottom's, divide. If the top is a constant and the bottom is a quadratic with no real zeros, complete the square. If the top is a constant multiple of the bottom's derivative, substitute instead, and neither rewrite is needed.

Why does a logarithm show up after long division?

The remainder sits over the divisor. When the divisor is linear, such as $x + 2$ , the piece $\frac{r}{x + 2}$ has antiderivative $r \ln | x + 2 | + C$ , where $r$ is the remainder.

Do I need the absolute value in

\ln | x + 2 |

In an indefinite integral, yes: $\ln | u |$ is an antiderivative of $\frac{1}{u}$ for negative $u$ as well as positive. Over an interval where $x + 2$ stays positive, such as $0 \leq x \leq 3$ , $\ln (x + 2)$ gives the same values.

Does completing the square only ever lead to an inverse tangent?

No. Under a square root, a completed square can match the derivative of the inverse sine: for $a > 0$ , $\int \frac{d u}{\sqrt{a^{2} - u^{2}}} = \arcsin (\frac{u}{a}) + C$ . For example, $4 x - x^{2} = 4 - (x - 2)^{2}$ , so $\int \frac{d x}{\sqrt{4 x - x^{2}}} = \arcsin (\frac{x - 2}{2}) + C$ .

How can I check an answer without a key?

Differentiate it. If the derivative of your antiderivative is the original fraction (after you combine terms over a common denominator), the answer is right, whatever form it is written in.

Course alignment, for teachers

AP Calculus AB topic 6.10, Unit 6: Integration and Accumulation of Change.

Skill 1.E: Apply appropriate mathematical rules or procedures, with and without technology.

This lesson maps the one essential knowledge statement of the topic to the part that teaches each one: 1 is checked by a scored question. The map is in the lesson file.