Selecting Techniques for
Antidifferentiation
Master the art of choosing the right integration method for any function: power rule, substitution, and long division strategies on the AP exam.
AP Calculus AB
What Is Antidifferentiation?
Antidifferentiation (or indefinite integration) is the process of finding a function \( F(x) \) whose derivative equals a given function \( f(x) \). In other words, if \( F'(x) = f(x) \), then:
The constant \( C \) appears because differentiation destroys constant terms: many functions can share the same derivative.
Why Technique Selection Matters
Unlike differentiation, which follows predictable rules, integration requires you to recognize patterns and choose the right approach. The AP exam tests this skill directly.
Speed
The correct technique turns a hard integral into a simple one in seconds.
Accuracy
Wrong methods lead to algebraic dead ends or incorrect answers.
Flexibility
Some integrals require rewriting before any technique applies.
The Three Core Techniques
1. Direct Formulas
Power rule, trig integrals, and exponentials: apply known antiderivatives immediately.
2. \(u\)-Substitution
Reverse the chain rule by identifying an inner function and its derivative.
3. Algebraic Rewriting
Simplify via long division, completing the square, or expanding before integrating.
Essential Antiderivative Formulas
| Function \( f(x) \) | Antiderivative \( F(x) + C \) |
|---|---|
| \( x^n \) (\(n \neq -1\)) | \( \dfrac{x^{n+1}}{n+1} \) |
| \( \dfrac{1}{x} \) | \( \ln|x| \) |
| \( e^x \) | \( e^x \) |
| \( \sin x \) | \( -\cos x \) |
| \( \cos x \) | \( \sin x \) |
| \( \sec^2 x \) | \( \tan x \) |
The \(u\)-Substitution Rule
This reverses the chain rule. Look for a composite function where the derivative of the inner function (or a constant multiple of it) is present as a factor.
Direct vs \(u\)-Substitution
| Direct Formula | \(u\)-Substitution | |
|---|---|---|
| When to use | Integrand matches a known formula exactly | Integrand is a composite function with its chain factor |
| Example | \( \int x^3\,dx \) | \( \int 2x\,e^{x^2}\,dx \) |
| Key signal | No composition: just a single "layer" | A function inside another, plus (a multiple of) its derivative |
| Difficulty | Immediate | Requires pattern recognition |
When Algebra Comes First
Sometimes no formula or substitution works as is. Rewrite the integrand algebraically before integrating:
1
Expand products: \( \int (x+1)^2\,dx \) becomes \( \int (x^2+2x+1)\,dx \).
2
Split fractions: \( \int \dfrac{x^2+3}{x}\,dx = \int \!\left(x + \dfrac{3}{x}\right)dx \).
3
Long division: When the degree of the numerator \( \geq \) the degree of the denominator, divide first.
4
Trig identities: Replace \( \sin^2 x \) with \( \dfrac{1 - \cos 2x}{2} \) before integrating.
The Decision Flowchart
How to Apply the Strategy
1
Scan the integrand. Does it match a basic formula? If yes, integrate directly.
2
Look for composition. Is there a function inside another? Identify a candidate for \(u\).
3
Check for the chain factor. Is \(g'(x)\) (or a constant multiple) present? If yes, use \(u\)-sub.
4
Try algebraic rewriting. Expand, split fractions, use long division, or apply an identity.
5
After rewriting, restart at Step 1. The simplified form often yields to a basic formula or \(u\)-sub.
Example 1: Direct Formula
1
Scan: Each term is a power of \( x \) (or a constant). Basic formulas apply directly.
2
Integrate term by term: \( 3 \cdot \dfrac{x^5}{5} - 2 \cdot \dfrac{x^2}{2} + 5x + C \)
3
Simplify: \( \dfrac{3x^5}{5} - x^2 + 5x + C \)
Example 2: \(u\)-Substitution
1
Spot the composition: \( \cos(\,\cdot\,) \) applied to the inner function \( x^3 \).
2
Set \( u = x^3 \), so \( du = 3x^2\,dx \). Notice \(6x^2\,dx = 2\,du\).
3
Substitute: \( \displaystyle\int 2\cos u\,du = 2\sin u + C \).
4
Back-substitute: \( 2\sin\!\left(x^3\right) + C \).
Example 3: Algebraic Rewriting
1
No direct formula fits, and no obvious \(u\)-sub works on the fraction as written.
2
Split the fraction: \( \dfrac{x^3}{x} + \dfrac{2}{x} = x^2 + \dfrac{2}{x} \).
3
Integrate term by term: \( \dfrac{x^3}{3} + 2\ln|x| + C \).
Example 4: Long Division First
1
Degree of numerator (2) \( \geq \) degree of denominator (1), so perform polynomial long division.
2
Divide: \( \dfrac{x^2+3x+5}{x+1} = x + 2 + \dfrac{3}{x+1} \).
3
Integrate: \( \dfrac{x^2}{2} + 2x + 3\ln|x+1| + C \).
Common Mistakes
✕ Forgetting \(+\,C\)
Every indefinite integral must include the constant of integration. The AP exam deducts for this.
✕ Wrong \(u\) choice
Picking \(u\) as the outer function instead of the inner one. If \(du\) doesn't appear, try a different \(u\).
✕ Missing the constant multiple
If \(du = 3x^2\,dx\) but you have \(6x^2\,dx\), you must account for the factor of 2 when substituting.
✕ Skipping simplification
Forcing \(u\)-sub on a fraction that should be split or divided first wastes time and creates errors.
✕ Not back-substituting
After integrating in terms of \(u\), always replace \(u\) with its expression in \(x\).
Technique Summary
| Technique | Signal / Cue | Example Integral |
|---|---|---|
| Direct formula | Single-layer: powers, trig, \(e^x\) | \( \int 5x^3\,dx \) |
| \(u\)-substitution | Composition + chain factor present | \( \int 2x\,e^{x^2}\,dx \) |
| Expand / split | Product or fraction that can be separated | \( \int \frac{x^2+1}{x}\,dx \) |
| Long division | Rational function: degree(num) \(\geq\) degree(den) | \( \int \frac{x^2}{x+1}\,dx \) |
| Trig identity | Powers of \(\sin x\)/\(\cos x\) that need reduction | \( \int \sin^2 x\,dx \) |
Question 1 of 5
Which technique is most efficient to evaluate \( \displaystyle\int 4x^3\cos\!\left(x^4\right)dx \)?
The integrand is \( \cos(x^4) \cdot 4x^3 \), a composite function with its chain factor. Setting \( u = x^4 \) gives \( du = 4x^3\,dx \), so the integral becomes \( \int \cos u\,du = \sin u + C = \sin(x^4) + C \).
Question 2 of 5
Evaluate \( \displaystyle\int \frac{x^2 + 4}{x}\,dx \).
Split the fraction: \( \frac{x^2}{x} + \frac{4}{x} = x + \frac{4}{x} \). Now integrate term by term: \( \int x\,dx + \int \frac{4}{x}\,dx = \frac{x^2}{2} + 4\ln|x| + C \).
Question 3 of 5
What is the first step to evaluate \( \displaystyle\int \frac{x^2 + x + 3}{x + 1}\,dx \)?
Since degree(numerator) = 2 \(\geq\) degree(denominator) = 1, you should perform long division first. Dividing gives \( x + \frac{3}{x+1} \), which integrates to \( \frac{x^2}{2} + 3\ln|x+1| + C \).
Question 4 of 5
Evaluate \( \displaystyle\int \frac{e^x}{1 + e^x}\,dx \).
Let \(u = 1 + e^x\), so \(du = e^x\,dx\). The integral becomes \(\int \frac{1}{u}\,du = \ln|u| + C = \ln|1 + e^x| + C\). The numerator is exactly the derivative of the denominator: classic \(u\)-sub.
Question 5 of 5
A student tries to evaluate \( \displaystyle\int \sin^2 x\,dx \) by setting \(u = \sin x\). What goes wrong?
If \(u = \sin x\), then \(du = \cos x\,dx\). The integrand \(\sin^2 x\,dx\) has no \(\cos x\) factor, so you cannot rewrite \(dx\) in terms of \(du\) without reintroducing \(x\). The correct approach is a trig identity: \(\sin^2 x = \frac{1 - \cos 2x}{2}\).