Antidifferentiation Formulas
Textbook Reference
| Primary source | OpenStax Calculus Volume 1, Section 4.10: “Antiderivatives” |
| Direct link | https://openstax.org/books/calculus-volume-1/pages/4-10-antiderivatives |
| Textbook used in class | Stewart, Calculus, Section 3.9: “Antiderivatives” (Table of Antidifferentiation Formulas, Example 2) |
Opening Scenario
Each antidifferentiation formula is a differentiation formula read in reverse. You already know that $\dfrac{d}{dx}(x^4) = 4x^3$. Reading it backward: “the antiderivative of $4x^3$ is $x^4$.” With a small adjustment for the coefficient, the antiderivative of $x^3$ is $x^4/4$.
This lesson assembles the most important antiderivative formulas from the derivatives you already know.
Quick Reference: Table of Antiderivatives
| Function $f(x)$ | Antiderivative $F(x)$ | Condition |
|---|---|---|
| $k$ (constant) | $kx$ | |
| $x^n$ | $\dfrac{x^{n+1}}{n+1}$ | $n \neq -1$ |
| $x^{-1} = \dfrac{1}{x}$ | $\ln|x|$ | $x \neq 0$ |
| $e^x$ | $e^x$ | |
| $a^x$ | $\dfrac{a^x}{\ln a}$ | $a > 0, a \neq 1$ |
| $\sin x$ | $-\cos x$ | |
| $\cos x$ | $\sin x$ | |
| $\sec^2 x$ | $\tan x$ | |
| $\sec x \tan x$ | $\sec x$ | |
| $\csc^2 x$ | $-\cot x$ | |
| $\dfrac{1}{\sqrt{1-x^2}}$ | $\arcsin x$ | $|x| < 1$ |
| $\dfrac{1}{1+x^2}$ | $\arctan x$ |
Always add $+ C$.
Rules:
- Constant multiple: $\displaystyle\int c f(x)\,dx = c\int f(x)\,dx$
- Sum/difference: $\displaystyle\int [f(x) \pm g(x)]\,dx = \int f(x)\,dx \pm \int g(x)\,dx$
Key Concepts
1. The Power Rule for Antiderivatives
Differentiation: $\dfrac{d}{dx}(x^{n+1}) = (n+1)x^n$.
Antidifferentiation: To undo the coefficient $(n+1)$, divide by it. $$\int x^n\,dx = \frac{x^{n+1}}{n+1} + C, \quad n \neq -1.$$
The exception $n = -1$ is critical: the power rule does not apply to $1/x$ because dividing by $n+1 = 0$ is undefined. The antiderivative of $1/x$ is $\ln|x| + C$ (from the derivative of $\ln|x|$).
Example 1. Find $\displaystyle\int x^5\,dx$ and $\displaystyle\int \sqrt{x}\,dx$.
For $x^5$: $n = 5$, antiderivative $= \dfrac{x^6}{6} + C$.
For $\sqrt{x} = x^{1/2}$: $n = 1/2$, antiderivative $= \dfrac{x^{3/2}}{3/2} + C = \dfrac{2}{3}x^{3/2} + C$.
Verify: $\dfrac{d}{dx}\left(\dfrac{2}{3}x^{3/2}\right) = \dfrac{2}{3} \cdot \dfrac{3}{2} x^{1/2} = x^{1/2}$. Confirmed.
Boxed answers: $\displaystyle\int x^5\,dx = \dfrac{x^6}{6} + C$; $\displaystyle\int \sqrt{x}\,dx = \dfrac{2}{3}x^{3/2} + C$.
2. Negative Powers
The power rule applies for negative $n$ as well (except $n = -1$).
Example 2. Find $\displaystyle\int \frac{1}{x^2}\,dx = \int x^{-2}\,dx$.
$n = -2 \neq -1$: antiderivative $= \dfrac{x^{-1}}{-1} + C = -\dfrac{1}{x} + C$.
Verify: $\dfrac{d}{dx}\!\left(-\dfrac{1}{x}\right) = \dfrac{1}{x^2}$. Confirmed.
3. Using Sum and Constant-Multiple Rules
Any polynomial can be antidifferentiated term by term.
Example 3. Find $\displaystyle\int (3x^4 - 5x^2 + x - 7)\,dx$.
Apply the power rule to each term and the constant rule to $-7$: $$= \frac{3x^5}{5} - \frac{5x^3}{3} + \frac{x^2}{2} - 7x + C.$$
One $C$ for the whole antiderivative. Each term produces its own arbitrary constant, but constants sum to a constant, so we write a single $C$ at the end.
Boxed answer: $\dfrac{3}{5}x^5 - \dfrac{5}{3}x^3 + \dfrac{x^2}{2} - 7x + C$.
4. Trigonometric Antiderivatives
These follow directly from derivative formulas.
| Derivative fact | Antiderivative |
|---|---|
| $(\sin x)' = \cos x$ | $\int \cos x\,dx = \sin x + C$ |
| $(-\cos x)' = \sin x$ | $\int \sin x\,dx = -\cos x + C$ |
| $(\tan x)' = \sec^2 x$ | $\int \sec^2 x\,dx = \tan x + C$ |
“$\int \sin x\,dx = \cos x + C$.” The sign is wrong. Differentiating $\cos x$ gives $-\sin x$, not $\sin x$. The antiderivative of $\sin x$ is $-\cos x + C$. Verify: $(-\cos x)' = \sin x$. Correct.
Example 4. Find $\displaystyle\int (2\cos x + 3\sec^2 x)\,dx$.
$= 2\sin x + 3\tan x + C$.
Verify: $\dfrac{d}{dx}(2\sin x + 3\tan x) = 2\cos x + 3\sec^2 x$. Confirmed.
5. Combining Formulas
Example 5. Find $\displaystyle\int \frac{x^3 - 2\sqrt{x} + 1}{x}\,dx$.
Rewrite by dividing each term by $x$: $$= \int \left(x^2 - 2x^{-1/2} + x^{-1}\right)\,dx.$$
Apply each formula:
- $\displaystyle\int x^2\,dx = \dfrac{x^3}{3}$.
- $\displaystyle\int -2x^{-1/2}\,dx = -2 \cdot \dfrac{x^{1/2}}{1/2} = -4\sqrt{x}$.
- $\displaystyle\int x^{-1}\,dx = \ln|x|$.
Boxed answer: $\dfrac{x^3}{3} - 4\sqrt{x} + \ln|x| + C$.
Recap. When the integrand is a fraction, dividing first (when possible) turns it into a sum of simpler terms that the power rule handles individually.
Common Errors Summary
| Error | Example | Correction |
|---|---|---|
| Using the power rule for $n = -1$ | $\int \frac{1}{x}\,dx = \frac{x^0}{0} + C$ | Division by zero; the antiderivative is $\ln|x| + C$ |
| Wrong sign for $\int \sin x\,dx$ | Writing $\cos x + C$ | The antiderivative is $-\cos x + C$; differentiate to check |
| Forgetting to divide by $n+1$ | Writing $\int x^4\,dx = x^5 + C$ | Should be $x^5/5 + C$; divide by the new exponent |
| Separate $C$ for each term | Writing $x^3/3 + C_1 - 5x + C_2$ | Use a single $C$; all the constants combine |
Leveled Practice
Level 1 -- Direct Application
Problem 1. Find $\displaystyle\int (4x^3 - 3x^2 + 2)\,dx$.
Show answer
$= x^4 - x^3 + 2x + C$.
Verify: $(x^4 - x^3 + 2x)' = 4x^3 - 3x^2 + 2$. Confirmed.
Problem 2. Find $\displaystyle\int (e^x - \sin x + \sec^2 x)\,dx$.
Show answer
$= e^x + \cos x + \tan x + C$.
Verify: $(e^x + \cos x + \tan x)' = e^x - \sin x + \sec^2 x$. Confirmed.
Level 2 -- Multiple Steps
Problem 3. Find $\displaystyle\int \frac{3 - 5\cos x}{\sin^2 x}\,dx$.
Show answer
Write $\dfrac{3}{\sin^2 x} - \dfrac{5\cos x}{\sin^2 x} = 3\csc^2 x - 5\dfrac{\cos x}{\sin^2 x}$.
For the first term: $\int 3\csc^2 x\,dx = -3\cot x$.
For the second: $\dfrac{\cos x}{\sin^2 x} = \dfrac{\cos x}{\sin x} \cdot \dfrac{1}{\sin x} = \cot x \cdot \csc x = \csc x \cot x$.
$\int -5\csc x \cot x\,dx = 5\csc x$ (since $(\csc x)' = -\csc x \cot x$, so $(-\csc x)' = \csc x \cot x$, giving $\int \csc x \cot x\,dx = -\csc x$).
Boxed answer: $-3\cot x + 5\csc x + C$.
Level 3 -- Deeper Problems
Problem 4. Explain why $\displaystyle\int x^{-1}\,dx = \ln|x| + C$ rather than $\ln x + C$.
Show answer
The natural logarithm $\ln x$ is only defined for $x > 0$. But $1/x$ is defined for $x < 0$ as well. For $x < 0$, $\ln|x| = \ln(-x)$, and $\dfrac{d}{dx}\ln(-x) = \dfrac{-1}{-x} = \dfrac{1}{x}$. So $\ln|x|$ is an antiderivative of $1/x$ on both $(-\infty, 0)$ and $(0, \infty)$.
Writing $\ln x + C$ excludes the case $x < 0$. Writing $\ln|x| + C$ covers both intervals of the domain of $1/x$.
Mastery Checklist
Mental Model
Each antidifferentiation formula is a derivative formula run in reverse. The power rule $\frac{d}{dx}(x^{n+1}) = (n+1)x^n$ becomes $\int x^n\,dx = \frac{x^{n+1}}{n+1}$: raise the power by one, divide by the new power.
The constant $C$ is always present because the derivative of any constant is zero. When you differentiate $F(x) + C$, the $C$ vanishes, giving back $f(x)$. So the $C$ is invisible in the derivative direction but essential in the antiderivative direction.
The fastest way to master the formulas is to practice deriving them from the derivative direction each time, rather than memorizing them directly. If you know every derivative formula, you already know every antiderivative formula.
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