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Reading Time: 6 min
Last Updated: March 19, 2026
Main Ideas: 4
Reading Time: 6 min
Last Updated: March 19, 2026
Main Ideas: 4

Topic 6.14 Notes – Selecting Techniques for Antidifferentiation

Verified for 2027 AP® Calculus AB Exam
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By this point, you know several techniques. The challenge now is recognizing which one fits a given integral. This is pattern recognition mixed with a little algebra cleanup when needed.

Choosing the Right Antiderivative Technique

When you see an integral, your brain should immediately sort it into one of these categories:

  • Basic algebraic rules
  • Exponentials or logarithms
  • Trigonometric functions
  • Inverse trig patterns
  • Composite functions → substitution
  • Rational functions → divide or complete the square

Most AB integrals fall cleanly into one of those. The hard part is recognizing the structure quickly.

The Core Patterns You Must Recognize

1. Power Rule and Basic Algebra

If it’s a polynomial or a sum of powers of x x , this is your easiest case.

∫xndx=xn+1n+1+C(n≠−1) \int x^n dx = \frac{x^{n+1}}{n+1} + C \quad (n \neq -1)

Other rules you already use automatically:

  • Constant multiple rule
  • Sum and difference rule

Special case you must treat differently:

∫1xdx=ln⁡∣x∣+C \int \frac{1}{x} dx = \ln|x| + C

Quick example:

∫(4x3−5x+2)dx=x4−52x2+2x+C \int (4x^3 - 5x + 2) dx = x^4 - \frac{5}{2}x^2 + 2x + C

Common quiz mistake: applying the power rule to x−1 x^{-1} . That one gives a logarithm, not a power.

2. Exponentials and Logarithms

Know these instantly:

  • ∫exdx=ex+C \int e^x dx = e^x + C
  • ∫axdx=axln⁡a+C \int a^x dx = \frac{a^x}{\ln a} + C
  • ∫1xdx=ln⁡∣x∣+C \int \frac{1}{x} dx = \ln|x| + C

If you see something like

∫33x+1dx \int \frac{3}{3x+1} dx

that denominator suggests a logarithm, but you need substitution because the derivative of 3x+1 3x+1 is involved.

Students often forget:

  • Absolute value in ln⁡∣x∣ \ln|x|
  • The 1ln⁡a \frac{1}{\ln a} factor for ax a^x

3. Trigonometric Integrals

These need to be automatic.

  • ∫sin⁡xdx=−cos⁡x+C \int \sin x dx = -\cos x + C
  • ∫cos⁡xdx=sin⁡x+C \int \cos x dx = \sin x + C
  • ∫tan⁡xdx=−ln⁡∣cos⁡x∣+C \int \tan x dx = -\ln|\cos x| + C
  • ∫sec⁡xdx=ln⁡∣sec⁡x+tan⁡x∣+C \int \sec x dx = \ln|\sec x + \tan x| + C

If you see something like

∫5cos⁡(5x)dx \int 5\cos(5x) dx

that inner function means substitution is needed.

Sign errors are extremely common with sine and cosine. Double-check mentally by differentiating your result.

4. Inverse Trigonometric Forms

These show up when the denominator matches a very specific pattern.

∫11+x2dx=tan⁡−1(x)+C \int \frac{1}{1+x^2} dx = \tan^{-1}(x) + C

∫11−x2dx=sin⁡−1(x)+C \int \frac{1}{\sqrt{1-x^2}} dx = \sin^{-1}(x) + C

You’ll also see variations like:

∫19+x2dx \int \frac{1}{9+x^2} dx

Rewrite 9 9 as 32 3^2 . The result becomes:

13tan⁡−1(x3)+C \frac{1}{3}\tan^{-1}\left(\frac{x}{3}\right) + C

If it’s close but not exact, you may need to complete the square first.

U-Substitution When It’s a Composite Function

This is reverse chain rule thinking.

You use substitution when:

  • Something is raised to a power
  • A trig or exponential function has an inner expression
  • A denominator looks like f(x) f(x) and the numerator looks like f′(x) f'(x)

Example:

∫2x(x2+4)5dx \int 2x(x^2+4)^5 dx

Let u=x2+4 u = x^2+4 . Then du=2xdx du = 2x dx .

The integral becomes:

∫u5du=u66+C \int u^5 du = \frac{u^6}{6} + C

Substitute back:

(x2+4)66+C \frac{(x^2+4)^6}{6} + C

Two common errors:

  • Leaving an x x behind after substitution
  • Trying substitution when the derivative of the inside is not present

On free-response questions, clean substitution work earns method points.

Rewriting Before Integrating

Sometimes the expression isn’t ready yet.

Long Division

If the degree of the numerator is greater than or equal to the denominator, divide first.

Example structure:

∫x2+1x−2dx \int \frac{x^2+1}{x-2} dx

After division, you’ll get a polynomial plus a rational remainder. Then integrate each piece separately. This often produces a logarithm from a 1x−a \frac{1}{x-a} term.

Completing the Square

Used when:

  • You see a quadratic in the denominator
  • You’re trying to match an inverse trig form

Example:

∫1x2+4x+8dx \int \frac{1}{x^2 + 4x + 8} dx

Rewrite x2+4x+8 x^2+4x+8 as:

(x+2)2+4 (x+2)^2 + 4

Now it matches the a2+x2 a^2 + x^2 structure and leads to an arctangent after substitution.

Students often forget to factor out a leading coefficient before completing the square. Watch for that.

Key Takeaways

Always simplify the integrand before choosing a method.
x−1 x^{-1} integrates to ln⁡∣x∣ \ln|x| , not a power rule expression.
Inverse trig forms must match 1+x2 1+x^2 or 1−x2 1-x^2 exactly after rewriting.
Substitution only works cleanly when the derivative of the inside function is present.
If the numerator’s degree is ≥ the denominator’s, divide first.
Mentally differentiating your answer is the fastest way to catch sign and scaling errors.

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Notes

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