Unit 6 · Topic 6.14 Beta

Choosing an Antidifferentiation Technique

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

Practice 1: Implementing Mathematical Processes

Question set for this topic

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=x2+1u = x^2 + 1, ∫x(x2+1)4dx\int x(x^2 + 1)^4\,dx becomes

  1. 12∫u4du\frac{1}{2}\int u^4\,du
  2. ∫u4du\int u^4\,du
  3. 2∫u4du2\int u^4\,du
  4. ∫xu4du\int x u^4\,du
Show the answer

du=2xdxdu = 2x\,dx, so xdx=12dux\,dx = \frac{1}{2}du.

  • Correct: 12∫u4du\frac{1}{2}\int u^4\,du:
  • ∫u4du\int u^4\,du:
  • 2∫u4du2\int u^4\,du:
  • ∫xu4du\int x u^4\,du:

2. x2+6x+10=x^2 + 6x + 10 =

  1. (x+3)2+1(x + 3)^2 + 1
  2. (x+6)2+10(x + 6)^2 + 10
  3. (x+3)2+10(x + 3)^2 + 10
  4. (x−3)2+1(x - 3)^2 + 1
Show the answer

Half of 6 is 3; 9+1=109 + 1 = 10.

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

Part 4 · See it

See it first

A flowchart of four 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. If none fits, rewrite and start again.
Ask the questions in order: basic form, substitution, long division, completing the square. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

  1. 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
  2. An inside function appears with its derivative as a factorsubstitute uu for the inside and change any limits
  3. A fraction has a top of degree at least the bottomdivide first, then integrate the pieces
  4. 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). ∫xex2dx\int x e^{x^2}\,dx yes; ∫ex2dx\int e^{x^2}\,dx no.
  • For fractions, compare: top is a multiple of the bottom's derivative (ln⁡\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: ∫6xx2+1dx\int \frac{6x}{x^2 + 1}\,dx=6arctan⁡x+C= 6\arctan x + C, since there is x2+1x^2 + 1 on the bottom.

What actually happens: The xx on top changes everything: it is a multiple of the bottom's derivative, so substitute u=x2+1u = x^2 + 1 and get 3ln⁡(x2+1)+C3\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
LabelIntegral
I∫xex2dx\int x e^{x^2}\,dx
II∫x3+2xdx\int \frac{x^3 + 2}{x}\,dx, x>0x > 0
III∫1x2+2x+2dx\int \frac{1}{x^2 + 2x + 2}\,dx
IV∫x2+1x−1dx\int \frac{x^2 + 1}{x - 1}\,dx, x>1x > 1

1. Which integral is set up for the substitution u=x2u = x^2?

  1. I
  2. II
  3. III
  4. IV
Show the answer

In I the inside function x2x^2 has derivative 2x2x, and xx is a factor: ∫xex2dx\int x e^{x^2}\,dx=12ex2+C= \frac{1}{2}e^{x^2} + C.

  • Correct: I: Correct: du=2xdxdu = 2x\,dx and the factor xx is present: 12∫eudu\frac{1}{2}\int e^u\,du.
  • II: Here dividing each term by xx gives basic forms, x2+2xx^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?

  1. II
  2. III
  3. I
  4. IV
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 xx: just divide each term of the top by xx. 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. x2+1x−1\frac{x^2 + 1}{x - 1}=x= x+1+ 1+2x−1+ \frac{2}{x - 1}.

3. Evaluate integral III, ∫1x2+2x+2dx\int \frac{1}{x^2 + 2x + 2}\,dx. Include +C+ C.

Type an expression for ∫1x2+2x+2dx=\int \frac{1}{x^2 + 2x + 2}\,dx =.

Show the answer

x2+2x+2x^2 + 2x + 2=(x+1)2+1= (x + 1)^2 + 1. With u=x+1u = x + 1: ∫duu2+1\int \frac{du}{u^2 + 1}=arctan⁡(x+1)+C= \arctan(x + 1) + C.

  • Answer: ∫1x2+2x+2dx=\int \frac{1}{x^2 + 2x + 2}\,dx = arctan⁡(x+1)+C\arctan(x + 1) + C

4. Evaluate integral IV, ∫x2+1x−1dx\int \frac{x^2 + 1}{x - 1}\,dx for x>1x > 1. Include +C+ C.

Type an expression for ∫x2+1x−1dx=\int \frac{x^2 + 1}{x - 1}\,dx =.

Show the answer

x2+1x−1\frac{x^2 + 1}{x - 1}=x= x+1+ 1+2x−1+ \frac{2}{x - 1}, so the integral is x22+x+2ln⁡|x−1|+C\frac{x^2}{2} + x + 2\ln|x - 1| + C.

  • Answer: ∫x2+1x−1dx=\int \frac{x^2 + 1}{x - 1}\,dx = x22+x+2ln⁡(|x−1|)+C\frac{x^{2}}{2} + x + 2\,\ln(\left|x - 1\right|) + C

5. Find ∫11+4x2dx\int \frac{1}{1 + 4x^2}\,dx.

  1. 12arctan⁡(2x)+C\frac{1}{2}\arctan(2x) + C
  2. arctan⁡(2x)+C\arctan(2x) + C
  3. 18xln⁡(1+4x2)+C\frac{1}{8x}\ln(1 + 4x^2) + C
  4. 14arctan⁡(2x)+C\frac{1}{4}\arctan(2x) + C
Show the answer

1+4x2=1+(2x)21 + 4x^2 = 1 + (2x)^2. Let u=2xu = 2x, du=2dxdu = 2\,dx: 12∫du1+u2\frac{1}{2}\int \frac{du}{1 + u^2}=12arctan⁡(2x)= \frac{1}{2}\arctan(2x)+C+ C.

  • Correct: 12arctan⁡(2x)+C\frac{1}{2}\arctan(2x) + C: Correct: u=2xu = 2x, du=2dxdu = 2\,dx, so 12∫du1+u2\frac{1}{2}\int \frac{du}{1 + u^2}.
  • arctan⁡(2x)+C\arctan(2x) + C: Its derivative is 21+4x2\frac{2}{1 + 4x^2}, twice the integrand.
  • 18xln⁡(1+4x2)+C\frac{1}{8x}\ln(1 + 4x^2) + C: The ln⁡\ln form needs 8x8x on top; and xx cannot be pulled out of an integral.
  • 14arctan⁡(2x)+C\frac{1}{4}\arctan(2x) + C: 1+4x2=1+(2x)21 + 4x^2 = 1 + (2x)^2; with u=2xu = 2x, dx=12dudx = \frac{1}{2}du, so the factor is 12\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?

  1. 2xx2+1\frac{2x}{x^2 + 1}
  2. ex2e^{x^2}
  3. xex2xe^{x^2}
  4. 1x2+1\frac{1}{x^2 + 1}
Show the answer

ex2e^{x^2} has antiderivatives (such as ∫0xet2dt\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.

  • 2xx2+1\frac{2x}{x^2 + 1}: Substitute u=x2+1u = x^2 + 1: ln⁡(x2+1)+C\ln(x^2 + 1) + C.
  • Correct: ex2e^{x^2}: Correct: no elementary function has derivative ex2e^{x^2}; without the factor xx, substitution does not apply.
  • xex2xe^{x^2}: Substitute u=x2u = x^2: 12ex2+C\frac{1}{2}e^{x^2} + C.
  • 1x2+1\frac{1}{x^2 + 1}: Its antiderivative is arctan⁡x+C\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.

Part 10 · Up next

What comes next

Part 11 · Connections

Connections