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Antidifferentiation Rules

Reference: Stewart §3.9  •  Chapter: 3  •  Section: 9

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Reading Differentiation Backwards

Every differentiation rule, when read in reverse, gives you an antidifferentiation rule. Know that $\frac{d}{dx}(x^3) = 3x^2$? Then you also know an antiderivative of $3x^2$ is $x^3$.

The key formulas in this section let you antidifferentiate polynomials, rational functions (with integer powers), and basic trig functions instantly, without computing any limits.

Think of it as building a vocabulary. The more differentiation formulas you've memorized, the more antiderivatives you can recognize on sight.

Prerequisite Map

This skillAntidifferentiation Rules

Quick Reference

Property Value
Concept Antiderivatives
Chapter 3.9
Difficulty Beginner
Time ~20 minutes

Key Concepts

The Power Rule for Antiderivatives

The derivative rule $\frac{d}{dx}(x^n) = nx^{n-1}$ reverses to:

$$\boxed{\int x^n\, dx = \frac{x^{n+1}}{n+1} + C \quad (n \neq -1)}$$

In words: "Add one to the power, divide by the new power."

Why does this work? Check by differentiating: $$\frac{d}{dx}\left(\frac{x^{n+1}}{n+1}\right) = \frac{(n+1)x^n}{n+1} = x^n \checkmark$$

Warning: This formula fails when $n = -1$ because you'd be dividing by zero. The antiderivative of $x^{-1} = 1/x$ is $\ln\vert x\vert + C$, which you'll learn later.

Constant Multiple Rule

$$\int cf(x)\,dx = c\int f(x)\,dx$$

Constants "pull out" of the antiderivative, just like with derivatives.

Sum and Difference Rules

$$\int [f(x) + g(x)]\,dx = \int f(x)\,dx + \int g(x)\,dx$$

$$\int [f(x) - g(x)]\,dx = \int f(x)\,dx - \int g(x)\,dx$$

You can antidifferentiate term by term.

Table of Basic Antiderivatives

Function $f(x)$ Antiderivative $F(x)$ Notes
$x^n$ $(n \neq -1)$ $\frac{x^{n+1}}{n+1} + C$ Power rule
$\sin x$ $-\cos x + C$ Derivative of $-\cos x$ is $\sin x$
$\cos x$ $\sin x + C$ Derivative of $\sin x$ is $\cos x$
$\sec^2 x$ $\tan x + C$ Derivative of $\tan x$ is $\sec^2 x$
$\sec x \tan x$ $\sec x + C$ Derivative of $\sec x$ is $\sec x \tan x$
$\csc^2 x$ $-\cot x + C$ Derivative of $-\cot x$ is $\csc^2 x$
$\csc x \cot x$ $-\csc x + C$ Derivative of $-\csc x$ is $\csc x \cot x$

Working with Negative and Fractional Exponents

Convert to exponential form before applying the power rule:

Expression Rewrite as Antiderivative
$\frac{1}{x^3}$ $x^{-3}$ $\frac{x^{-2}}{-2} + C = -\frac{1}{2x^2} + C$
$\sqrt{x}$ $x^{1/2}$ $\frac{x^{3/2}}{3/2} + C = \frac{2}{3}x^{3/2} + C$
$\frac{1}{\sqrt{x}}$ $x^{-1/2}$ $\frac{x^{1/2}}{1/2} + C = 2\sqrt{x} + C$
$\sqrt[3]{x^2}$ $x^{2/3}$ $\frac{x^{5/3}}{5/3} + C = \frac{3}{5}x^{5/3} + C$

Combining Rules: Polynomials

For a polynomial, antidifferentiate term by term:

$$\int (4x^3 - 5x^2 + 2x - 7)\,dx = x^4 - \frac{5x^3}{3} + x^2 - 7x + C$$

Key: Only ONE constant $C$ at the end, not one per term!

Practice Problems

Level 1 Direct Power Rule

Find the general antiderivative of $f(x) = x^5$.

Thought Process

Apply the power rule: add 1 to the exponent (getting 6), then divide by the new exponent.

Show Answer

Using the power rule with $n = 5$:

$$\int x^5\,dx = \frac{x^{5+1}}{5+1} + C = \frac{x^6}{6} + C$$

Level 2 Root Function

Find the general antiderivative of $f(x) = \sqrt{x}$.

Thought Process

First convert: $\sqrt{x} = x^{1/2}$. Then apply the power rule. Be careful with fraction arithmetic: $1/2 + 1 = 3/2$.

Show Answer

Rewrite: $\sqrt{x} = x^{1/2}$

Apply the power rule: $$\int x^{1/2}\,dx = \frac{x^{1/2 + 1}}{1/2 + 1} + C = \frac{x^{3/2}}{3/2} + C$$

Simplify (dividing by $3/2$ means multiplying by $2/3$): $$= \frac{2}{3}x^{3/2} + C = \frac{2}{3}x\sqrt{x} + C$$

Level 3 Polynomial

Find the general antiderivative of $g(x) = 6x^2 - 4x + 9$.

Thought Process

Use the sum rule to handle term by term. Apply the constant multiple rule to pull out coefficients. Remember: constant term $9$ antidifferentiates to $9x$.

Show Answer

Antidifferentiate term by term:

$$\int (6x^2 - 4x + 9)\,dx = 6 \cdot \frac{x^3}{3} - 4 \cdot \frac{x^2}{2} + 9x + C$$

$$= 2x^3 - 2x^2 + 9x + C$$

Check: $\frac{d}{dx}(2x^3 - 2x^2 + 9x + C) = 6x^2 - 4x + 9$ ✓

Level 4 Simplify Before Integrating

Find the general antiderivative of

$$h(x) = \frac{3x^4 - 2\sqrt{x}}{x}$$

Thought Process

First simplify by dividing each term by $x$. Rewrite roots as fractional exponents. Then apply the power rule to each term.

Show Answer

First, simplify by dividing each term by $x$:

$$h(x) = \frac{3x^4}{x} - \frac{2\sqrt{x}}{x} = 3x^3 - \frac{2x^{1/2}}{x} = 3x^3 - 2x^{-1/2}$$

Now antidifferentiate:

$$\int (3x^3 - 2x^{-1/2})\,dx = 3 \cdot \frac{x^4}{4} - 2 \cdot \frac{x^{1/2}}{1/2} + C$$

$$= \frac{3x^4}{4} - 4x^{1/2} + C = \frac{3x^4}{4} - 4\sqrt{x} + C$$

Level 5 Trig and Algebraic Combination

Find all functions $f$ such that

$$f'(x) = 5\sin x + \frac{3}{x^4} - 2\sec^2 x + 7$$

Thought Process

Handle each term separately:

  • $\sin x$ antidifferentiates to $-\cos x$
  • $3/x^4 = 3x^{-4}$ uses the power rule
  • $\sec^2 x$ antidifferentiates to $\tan x$
  • Constant 7 antidifferentiates to $7x$
Show Answer

Antidifferentiate term by term:

Term 1: $\int 5\sin x\,dx = -5\cos x$

Term 2: $\int 3x^{-4}\,dx = 3 \cdot \frac{x^{-3}}{-3} = -x^{-3} = -\frac{1}{x^3}$

Term 3: $\int -2\sec^2 x\,dx = -2\tan x$

Term 4: $\int 7\,dx = 7x$

Combining: $$f(x) = -5\cos x - \frac{1}{x^3} - 2\tan x + 7x + C$$

Verify: Differentiate to check you get back $f'(x)$. ✓

CCI-Style Conceptual Questions

Conceptual Pattern Recognition

Without computing, predict: is the antiderivative of $x^{99}$ a polynomial of degree 98, 99, or 100?

Thought Process

The power rule says: add 1 to the exponent. What does this tell you about the degree?

Show Answer

Degree 100.

By the power rule, the antiderivative of $x^{99}$ is $\frac{x^{100}}{100} + C$, which is degree 100.

In general: antidifferentiating increases the degree by 1, while differentiating decreases it by 1.

Conceptual Why n ≠ -1?

Explain in your own words why the formula $\int x^n\,dx = \frac{x^{n+1}}{n+1} + C$ cannot be used when $n = -1$.

Thought Process

What happens in the formula when $n = -1$? What mathematical operation becomes impossible?

Show Answer

When $n = -1$: $$\frac{x^{-1+1}}{-1+1} = \frac{x^0}{0} = \frac{1}{0}$$

Division by zero is undefined, so the formula breaks down.

The antiderivative of $x^{-1} = \frac{1}{x}$ exists, but it's a completely different function: $\ln\vert x\vert + C$ (which you'll learn in the chapter on transcendental functions).

This is one of the most important special cases in calculus!

Conceptual Derivative-Antiderivative Pairs

True or False: If $\frac{d}{dx}[F(x)] = f(x)$, then $\frac{d}{dx}[f(x)] = F(x)$.

Thought Process

This claims that differentiation and antidifferentiation are "symmetric." Test with a specific example, like $F(x) = x^2$ and $f(x) = 2x$.

Show Answer

False.

Counterexample: Let $F(x) = x^2$, so $f(x) = F'(x) = 2x$.

But $\frac{d}{dx}[f(x)] = \frac{d}{dx}[2x] = 2$, which is NOT equal to $F(x) = x^2$.

The correct relationship is:

  • $F$ is an antiderivative of $f$ means $F' = f$
  • The derivative of $f$ is $f'$, not $F$

Differentiation and antidifferentiation are inverse processes, but that doesn't mean applying one then the other gives you back what you started with. It only works one direction at a time.

Mastery Checklist

Mental Model

Antidifferentiation is "undoing" differentiation:

Differentiation (Power Rule) Antidifferentiation (Reverse Power Rule)
Multiply by the power Divide by the new power
Subtract 1 from the power Add 1 to the power

Think of it as watching a movie in reverse: every step of differentiation has a corresponding "undo" step in antidifferentiation.

But remember: antidifferentiation loses information (the "+C"), so you can't fully recover the original function without additional information.


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Antiderivative Definition Skills Index Initial Value Problems

Last updated: 2026-01-22