Unit 6 · Topic 6.14

Choosing an Antidifferentiation Technique

To choose a technique, ask in order: is it a basic form after rewriting?

Practice 1: Implementing Mathematical Processes

Part 1 · Hook

Why this matters

On the exam, nobody tells you which method to use. A page of integrals looks like a page of puzzles until you learn to ask a few questions in the same order every time. Then most integrals sort themselves in seconds, and the rest yield to one more rewrite.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. With $u = x^2 + 1$ , $\int x(x^2 + 1)^4\,dx$ becomes

$\frac{1}{2}\int u^4\,du$
$\int u^4\,du$
$2\int u^4\,du$
$\int x u^4\,du$

Show the answer

$du = 2x\,dx$ , so $x\,dx = \frac{1}{2}du$ .

Correct: $\frac{1}{2}\int u^4\,du$ :
$\int u^4\,du$ :
$2\int u^4\,du$ :
$\int x u^4\,du$ :

2. $x^2 + 6x + 10 =$

$(x + 3)^2 + 1$
$(x + 6)^2 + 10$
$(x + 3)^2 + 10$
$(x - 3)^2 + 1$

Show the answer

Half of 6 is 3; $9 + 1 = 10$ .

Correct: $(x + 3)^2 + 1$ :
$(x + 6)^2 + 10$ :
$(x + 3)^2 + 10$ :
$(x - 3)^2 + 1$ :

Part 4 · See it

See it first

$A flowchart of six questions in order, each with a yes arrow to a technique: basic form after rewriting? basic rules. Inside function whose derivative is a factor? substitution, with new limits. Rational with top degree at least bottom degree? long division first. Quadratic on the bottom with no real zeros? complete the square, then arctan. BC only: a product with a factor that gets simpler when differentiated? integration by parts. BC only: bottom with distinct linear factors? linear partial fractions. If none fits, rewrite and start again.$

Ask the questions in order: basic form, substitution, long division, completing the square, then the two BC questions: integration by parts and linear partial fractions. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

The integrand is a sum of powers, exponentials or trig functionsuse the basic rules, after rewriting roots as powers and splitting fractions over one term
An inside function appears with its derivative as a factorsubstitute $u$ for the inside and change any limits
A fraction has a top of degree at least the bottomdivide first, then integrate the pieces
A constant sits over a quadratic with no real zeroscomplete the square and use the arctan form

Part 6 · Key ideas

Key ideas

Rewrite first: roots as powers, products expanded, fractions over one term split.
Substitution needs the inside's derivative as a factor (up to a constant). $\int x e^{x^2}\,dx$ yes; $\int e^{x^2}\,dx$ no.
For fractions, compare: top is a multiple of the bottom's derivative ( $\ln$ ); top degree too high (divide); constant over an irreducible quadratic (arctan).
Always check by differentiating. If no method works, the definite integral may be a calculator question.

Part 7 · Misconception

A common mistake

The wrong idea: $\int \frac{6x}{x^2 + 1}\,dx$ $= 6\arctan x + C$ , since there is $x^2 + 1$ on the bottom.

What actually happens: The $x$ on top changes everything: it is a multiple of the bottom's derivative, so substitute $u = x^2 + 1$ and get $3\ln(x^2 + 1) + C$ . Arctan is for a constant on top.

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Table

Four integrals, four approaches

Each integral below needs a different first step.

Four indefinite integrals
Label	Integral
I	$\int x e^{x^2}\,dx$
II	$\int \frac{x^3 + 2}{x}\,dx$ , $x > 0$
III	$\int \frac{1}{x^2 + 2x + 2}\,dx$
IV	$\int \frac{x^2 + 1}{x - 1}\,dx$ , $x > 1$

1. Which integral is set up for the substitution $u = x^2$ ?

Show the answer

In I the inside function $x^2$ has derivative $2x$ , and $x$ is a factor: $\int x e^{x^2}\,dx$ $= \frac{1}{2}e^{x^2} + C$ .

Correct: I: Correct: $du = 2x\,dx$ and the factor $x$ is present: $\frac{1}{2}\int e^u\,du$ .
II: Here dividing each term by $x$ gives basic forms, $x^2 + \frac{2}{x}$ ; no inside function is needed.
III: Its bottom is a quadratic with no real zeros and the top is constant: complete the square.
IV: The top's degree is at least the bottom's: long division comes first.

2. Which integral should start with polynomial long division?

Show the answer

Long division is for a fraction whose top has degree at least the bottom's degree, with a bottom of more than one term: IV.

II: The bottom is the single term $x$ : just divide each term of the top by $x$ . No long division is needed.
III: The top has degree 0, less than the bottom's 2; complete the square instead.
I: There is no fraction to divide; substitute u = x squared.
Correct: IV: Correct: degree 2 over degree 1. $\frac{x^2 + 1}{x - 1}$ $= x$ $+ 1$ $+ \frac{2}{x - 1}$ .

3. Evaluate integral III, $\int \frac{1}{x^2 + 2x + 2}\,dx$ . Include $+ C$ .

Type an expression for $\int \frac{1}{x^2 + 2x + 2}\,dx =$ .

Show the answer

$x^2 + 2x + 2$ $= (x + 1)^2 + 1$ . With $u = x + 1$ : $\int \frac{du}{u^2 + 1}$ $= \arctan(x + 1) + C$ .

Answer: $\int \frac{1}{x^2 + 2x + 2}\,dx =$ $\arctan(x + 1) + C$

4. Evaluate integral IV, $\int \frac{x^2 + 1}{x - 1}\,dx$ for $x > 1$ . Include $+ C$ .

Type an expression for $\int \frac{x^2 + 1}{x - 1}\,dx =$ .

Show the answer

$\frac{x^2 + 1}{x - 1}$ $= x$ $+ 1$ $+ \frac{2}{x - 1}$ , so the integral is $\frac{x^2}{2} + x + 2\ln|x - 1| + C$ .

Answer: $\int \frac{x^2 + 1}{x - 1}\,dx =$ $\frac{x^{2}}{2} + x + 2\,\ln(\left|x - 1\right|) + C$

5. Find $\int \frac{1}{1 + 4x^2}\,dx$ .

$\frac{1}{2}\arctan(2x) + C$
$\arctan(2x) + C$
$\frac{1}{8x}\ln(1 + 4x^2) + C$
$\frac{1}{4}\arctan(2x) + C$

Show the answer

$1 + 4x^2 = 1 + (2x)^2$ . Let $u = 2x$ , $du = 2\,dx$ : $\frac{1}{2}\int \frac{du}{1 + u^2}$ $= \frac{1}{2}\arctan(2x)$ $+ C$ .

Correct: $\frac{1}{2}\arctan(2x) + C$ : Correct: $u = 2x$ , $du = 2\,dx$ , so $\frac{1}{2}\int \frac{du}{1 + u^2}$ .
$\arctan(2x) + C$ : Its derivative is $\frac{2}{1 + 4x^2}$ , twice the integrand.
$\frac{1}{8x}\ln(1 + 4x^2) + C$ : The $\ln$ form needs $8x$ on top; and $x$ cannot be pulled out of an integral.
$\frac{1}{4}\arctan(2x) + C$ : $1 + 4x^2 = 1 + (2x)^2$ ; with $u = 2x$ , $dx = \frac{1}{2}du$ , so the factor is $\frac{1}{2}$ .

6. Which of these has no antiderivative that can be written with the functions of this course, so a definite integral of it is found with a calculator?

$\frac{2x}{x^2 + 1}$
$e^{x^2}$
$xe^{x^2}$
$\frac{1}{x^2 + 1}$

Show the answer

$e^{x^2}$ has antiderivatives (such as $\int_0^x e^{t^2}\,dt$ ), but none built from polynomials, roots, exponentials, logs and trig functions. Definite integrals of it are found numerically.

$\frac{2x}{x^2 + 1}$ : Substitute $u = x^2 + 1$ : $\ln(x^2 + 1) + C$ .
Correct: $e^{x^2}$ : Correct: no elementary function has derivative $e^{x^2}$ ; without the factor $x$ , substitution does not apply.
$xe^{x^2}$ : Substitute $u = x^2$ : $\frac{1}{2}e^{x^2} + C$ .
$\frac{1}{x^2 + 1}$ : Its antiderivative is $\arctan x + C$ .

Part 9 · Summary

Summary

To choose a technique, ask in order: is it a basic form after rewriting? Is there an inside function with its derivative as a factor? Is it a fraction to divide? Is it a constant over an irreducible quadratic? Then integrate, check by differentiating, and change the limits whenever you substitute in a definite integral. On the BC exam, two more questions follow: is it a product with a factor that simplifies when differentiated (integration by parts)? Does the bottom factor into distinct linear factors (linear partial fractions)?

Part 10 · Up next

What comes next

Part 11 · Connections

Connections

Back: long division and completing the square (6.10)

Question set for this topic