Antiderivatives and Indefinite Integrals
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 |
| Supplementary | OpenStax Calculus Volume 1, Section 5.5: “Substitution” |
| Supplementary link | https://openstax.org/books/calculus-volume-1/pages/5-5-substitution |
Both sources are free and openly licensed.
Key idea
A differential equation asks you to find a function when you are told something about its rate of change. The most basic version of that question is: “I know the derivative. What was the original function?” That question is exactly what an antiderivative answers. Every method in a differential equations course rests on this one skill, so it belongs at the very start.
There is one feature of this question that surprises students, and it is worth meeting head on. Differentiation throws away constants. The derivative of $x^2 + 7$ and the derivative of $x^2 - 4$ are both $2x$, because the derivative of any constant is zero. Running the process backward, then, cannot recover a single function. It recovers a whole family of functions that differ only by a constant. That lost constant is the reason every indefinite integral carries a $+ C$, and it is the same constant an initial condition will later pin down.
So when you see $+ C$, do not treat it as a ritual mark. It is an honest statement that the original information was not enough to name one function, and it is the seam where a differential equation and a starting value meet.
Prerequisite Check
Before this lesson, make sure you can do all of the following:
If the chain rule is shaky, review The Chain Rule first.
Quick Reference
Definition. A function $F$ is an antiderivative of $f$ on an interval if \[ F'(x) = f(x) \quad \text{for every } x \text{ in that interval.} \]
The family. If $F$ is one antiderivative of $f$, then every antiderivative has the form $F(x) + C$ for a constant $C$. The indefinite integral collects the whole family: \[ \int f(x)\, dx = F(x) + C. \]
The core rules (each is a derivative rule read backward; $C$ is an arbitrary constant):
| $f(x)$ | $\displaystyle \int f(x)\, dx$ | Condition |
|---|---|---|
| $x^n$ | $\dfrac{x^{n+1}}{n+1} + C$ | $n \neq -1$ |
| $\dfrac{1}{x}$ | $\ln\lvert x\rvert + C$ | $x \neq 0$ |
| $e^{x}$ | $e^{x} + C$ | |
| $e^{kx}$ | $\dfrac{1}{k}e^{kx} + C$ | $k \neq 0$ |
| $\cos x$ | $\sin x + C$ | |
| $\sin x$ | $-\cos x + C$ |
Linearity. Constants pull out and sums split: \[ \int \left[a\, f(x) + b\, g(x)\right] dx = a \int f(x)\, dx + b \int g(x)\, dx. \]
Key Concepts
1. What an Antiderivative Is
An antiderivative of $f$ is any function $F$ whose derivative is $f$. The word says exactly what it means: it is the reverse of taking a derivative.
To check a claimed antiderivative, you never need new tools. You differentiate your answer and see whether you get back the original function.
Example 1. Show that $F(x) = x^3$ is an antiderivative of $f(x) = 3x^2$.
Differentiate $F$: \[ F'(x) = \frac{d}{dx}\left(x^3\right) = 3x^2 = f(x). \] Since $F'(x) = f(x)$, the function $F$ is an antiderivative of $f$.
Important: an antiderivative is verified by differentiating, not by integrating again. The fastest way to catch a mistake in any integral is to differentiate your answer. If it does not return the integrand, the answer is wrong. This check is free and you should use it every time.
2. Why Every Antiderivative Carries a Constant
The derivative of a constant is zero. So if $F'(x) = f(x)$, then for any constant $C$, \[ \frac{d}{dx}\left[F(x) + C\right] = F'(x) + 0 = f(x). \] Every function of the form $F(x) + C$ is also an antiderivative of $f$. It is a theorem of calculus that there are no others on an interval: any two antiderivatives of the same function differ by a constant.
Example 2. Find all antiderivatives of $f(x) = 2x$.
One antiderivative is $x^2$, because $\dfrac{d}{dx}\left(x^2\right) = 2x$. Every antiderivative therefore has the form \[ \int 2x\, dx = x^2 + C. \] The graphs $y = x^2$, $y = x^2 + 1$, and $y = x^2 - 3$ are vertical shifts of one another. They all have the same slope $2x$ at each $x$, which is why they share a derivative.
3. The Power Rule, Run Backward
To undo $\dfrac{d}{dx}\left(x^{n+1}\right) = (n+1)x^n$, raise the exponent by one and divide by the new exponent: \[ \int x^n\, dx = \frac{x^{n+1}}{n+1} + C \qquad (n \neq -1). \]
The condition $n \neq -1$ is not optional. If $n = -1$, the formula would divide by zero. The case $n = -1$, meaning $\int \frac{1}{x}\, dx$, has its own rule (Concept 4).
Example 3. Compute $\displaystyle \int x^4\, dx$.
Raise the exponent to $5$ and divide by $5$: \[ \int x^4\, dx = \frac{x^5}{5} + C. \] Check: $\dfrac{d}{dx}\left(\frac{x^5}{5}\right) = \frac{5x^4}{5} = x^4$. Correct.
Example 4. Compute $\displaystyle \int \frac{1}{x^3}\, dx$.
Rewrite with a negative exponent first: $\frac{1}{x^3} = x^{-3}$. Then $n = -3$, and $n + 1 = -2$: \[ \int x^{-3}\, dx = \frac{x^{-2}}{-2} + C = -\frac{1}{2x^2} + C. \] Check: $\dfrac{d}{dx}\left(-\frac{1}{2}x^{-2}\right) = -\frac{1}{2}(-2)x^{-3} = x^{-3}$. Correct.
forgetting to rewrite roots and reciprocals as powers. Expressions like $\sqrt{x}$ and $\frac{1}{x^2}$ must be written as $x^{1/2}$ and $x^{-2}$ before the power rule applies. Students who try to integrate $\sqrt{x}$ without rewriting often guess; rewriting makes the rule mechanical.
4. The Logarithm Case and the Exponential Rules
The one power the power rule cannot handle is $x^{-1}$. Its antiderivative comes from the derivative of the natural logarithm: \[ \int \frac{1}{x}\, dx = \ln\lvert x\rvert + C. \] The absolute value matters. The function $\frac{1}{x}$ is defined for negative $x$ as well, and $\ln\lvert x\rvert$ extends the antiderivative to that side correctly.
For exponentials, the derivative of $e^x$ is itself, so \[ \int e^x\, dx = e^x + C. \] When the exponent has a constant multiplier, the chain rule produces an extra factor of $k$ on the way down, so you divide by $k$ on the way up: \[ \int e^{kx}\, dx = \frac{1}{k}e^{kx} + C \qquad (k \neq 0). \]
Example 5. Compute $\displaystyle \int e^{5x}\, dx$.
Here $k = 5$, so divide by $5$: \[ \int e^{5x}\, dx = \frac{1}{5}e^{5x} + C. \] Check: $\dfrac{d}{dx}\left(\frac{1}{5}e^{5x}\right) = \frac{1}{5}\cdot 5 e^{5x} = e^{5x}$. Correct.
The absolute value in $\ln\lvert x\rvert$ is not decoration. Writing $\int \frac{1}{x}\, dx = \ln x + C$ silently restricts the answer to $x > 0$. Differential equations frequently produce $\int \frac{1}{y}\, dy$ where $y$ can be negative, so the absolute value carries real information. Keep it.
5. Linearity: Splitting Sums and Pulling Out Constants
Differentiation is linear, so integration is too. A constant multiplier moves through the integral sign, and a sum of terms integrates term by term: \[ \int \left[a\, f(x) + b\, g(x)\right] dx = a \int f(x)\, dx + b \int g(x)\, dx. \]
Example 6. Compute $\displaystyle \int \left(6x^2 - 4x + 3\right) dx$.
Integrate each term: \[ \int 6x^2\, dx - \int 4x\, dx + \int 3\, dx = 6\cdot\frac{x^3}{3} - 4\cdot\frac{x^2}{2} + 3x + C = 2x^3 - 2x^2 + 3x + C. \] Note that only one constant $C$ is written, not one per term. The separate constants combine into a single arbitrary constant.
writing a separate $+C$ for each term. A single arbitrary constant covers the whole expression. Writing $+ C_1 + C_2 + C_3$ is not wrong in spirit, but it is clutter, since the sum of arbitrary constants is just one arbitrary constant.
6. Substitution: Undoing the Chain Rule
When the integrand is a composite times the derivative of the inside, the chain rule run backward is called substitution (or $u$-substitution). If the integrand has the form $f(g(x))\, g'(x)$, set $u = g(x)$, so $du = g'(x)\, dx$, and the integral becomes $\int f(u)\, du$.
Example 7. Compute $\displaystyle \int 2x\, e^{x^2}\, dx$.
The inside function is $x^2$, and its derivative $2x$ is present as a factor. Let $u = x^2$, so $du = 2x\, dx$. The integral becomes \[ \int e^{u}\, du = e^{u} + C = e^{x^2} + C. \] Check: $\dfrac{d}{dx}\left(e^{x^2}\right) = e^{x^2}\cdot 2x = 2x\, e^{x^2}$. Correct.
Substitution is the single most important integration technique for a differential equations course, because separating variables produces integrals of exactly this shape.
Common Errors Summary
| Error | Example | Correction |
|---|---|---|
| Forgetting the constant of integration | $\int 2x\, dx = x^2$ | $\int 2x\, dx = x^2 + C$ |
| Applying the power rule to $x^{-1}$ | $\int \frac{1}{x}\, dx = \frac{x^0}{0}$ | $\int \frac{1}{x}\, dx = \ln\lvert x\rvert + C$ |
| Dropping the absolute value | $\int \frac{1}{x}\, dx = \ln x + C$ | Keep $\ln\lvert x\rvert + C$ |
| Not rewriting roots as powers | integrating $\sqrt{x}$ directly | Write $x^{1/2}$, then apply the power rule |
| Missing the $\frac{1}{k}$ on $e^{kx}$ | $\int e^{3x}\, dx = e^{3x} + C$ | $\int e^{3x}\, dx = \frac{1}{3}e^{3x} + C$ |
| Not checking by differentiating | accepting an answer untested | Differentiate the result; it must return the integrand |
Leveled Practice
Level 1 -- Direct Application
Problem 1. Compute $\displaystyle \int x^7\, dx$.
Show answer
Raise the exponent to $8$ and divide by $8$: \[ \int x^7\, dx = \frac{x^8}{8} + C. \] Check: $\dfrac{d}{dx}\left(\frac{x^8}{8}\right) = \frac{8x^7}{8} = x^7$. $\checkmark$
Problem 2. Compute $\displaystyle \int \left(4x^3 + 2\right) dx$.
Show answer
Term by term: \[ 4\cdot\frac{x^4}{4} + 2x + C = x^4 + 2x + C. \] Check: $\dfrac{d}{dx}\left(x^4 + 2x\right) = 4x^3 + 2$. $\checkmark$
Problem 3. Compute $\displaystyle \int e^{-2x}\, dx$.
Show answer
Here $k = -2$, so divide by $-2$: \[ \int e^{-2x}\, dx = -\frac{1}{2}e^{-2x} + C. \] Check: $\dfrac{d}{dx}\left(-\frac{1}{2}e^{-2x}\right) = -\frac{1}{2}\cdot(-2)e^{-2x} = e^{-2x}$. $\checkmark$
Level 2 -- Rewriting First
Problem 4. Compute $\displaystyle \int \sqrt{x}\, dx$.
Show answer
Rewrite $\sqrt{x} = x^{1/2}$. Then $n = \frac{1}{2}$ and $n + 1 = \frac{3}{2}$: \[ \int x^{1/2}\, dx = \frac{x^{3/2}}{3/2} + C = \frac{2}{3}x^{3/2} + C. \] Check: $\dfrac{d}{dx}\left(\frac{2}{3}x^{3/2}\right) = \frac{2}{3}\cdot\frac{3}{2}x^{1/2} = x^{1/2}$. $\checkmark$
Problem 5. Compute $\displaystyle \int \frac{3}{x}\, dx$.
Show answer
Pull out the constant: \[ 3 \int \frac{1}{x}\, dx = 3\ln\lvert x\rvert + C. \] Check: $\dfrac{d}{dx}\left(3\ln\lvert x\rvert\right) = 3\cdot\frac{1}{x} = \frac{3}{x}$. $\checkmark$
Problem 6. Compute $\displaystyle \int \frac{x^2 + 1}{x}\, dx$.
Show answer
Split the fraction before integrating: \[ \frac{x^2 + 1}{x} = x + \frac{1}{x}. \] \[ \int \left(x + \frac{1}{x}\right) dx = \frac{x^2}{2} + \ln\lvert x\rvert + C. \] Check: $\dfrac{d}{dx}\left(\frac{x^2}{2} + \ln\lvert x\rvert\right) = x + \frac{1}{x}$. $\checkmark$
Level 3 -- Substitution and Connections
Problem 7. Compute $\displaystyle \int 3x^2\, e^{x^3}\, dx$.
Show answer
Let $u = x^3$, so $du = 3x^2\, dx$. The integral becomes \[ \int e^{u}\, du = e^{u} + C = e^{x^3} + C. \] Check: $\dfrac{d}{dx}\left(e^{x^3}\right) = e^{x^3}\cdot 3x^2 = 3x^2 e^{x^3}$. $\checkmark$
Problem 8. Solve the simplest differential equation: find every function $y$ with $\dfrac{dy}{dx} = 6x$, then find the one with $y(0) = 5$.
Show answer
The family of solutions is the indefinite integral: \[ y = \int 6x\, dx = 3x^2 + C. \] Apply the initial condition $y(0) = 5$: substitute $x = 0$ and $y = 5$: \[ 5 = 3(0)^2 + C = C, \qquad \text{so } C = 5. \] The one solution that passes through $(0, 5)$ is $y = 3x^2 + 5$.
This is the entire arc of a differential equations problem in miniature: integrate to get a family with a constant, then use a starting value to name the constant.
Problem 9. Compute $\displaystyle \int \frac{1}{2y}\, dy$, the integral that appears when you separate $\dfrac{dy}{dx} = 2xy$.
Show answer
Pull out the constant $\frac{1}{2}$: \[ \int \frac{1}{2y}\, dy = \frac{1}{2}\int \frac{1}{y}\, dy = \frac{1}{2}\ln\lvert y\rvert + C. \] The absolute value is kept because $y$ may be negative. This is exactly the kind of integral the separable-equations method produces, which is why the $\ln\lvert y\rvert$ rule is worth securing now.
Mastery Checklist
You have mastered this skill when you can do all of the following without referring to notes:
Mental Model
Think of differentiation and antidifferentiation as a one-way door with a small leak.
Going forward, from a function to its derivative, is clean: every function has exactly one derivative. Going backward leaks information, because the constant term has already been flattened to zero and cannot be recovered. So the reverse process returns not one function but a stack of parallel curves, each a vertical shift of the others. That stack is the family $F(x) + C$.
An initial condition is the single nail that fixes the stack in place. Before you know a starting value, you have a whole family of curves with the right slope everywhere. After you know one point the curve must pass through, exactly one curve in the stack qualifies, and $C$ is determined.
Holding this picture makes the entire structure of a first differential equations course feel inevitable: a differential equation describes slopes, integration recovers a family of curves, and a starting value selects the one curve that matches reality.
Connections
Within MATH347 (Differential Equations)
- Introduction to differential equations: The simplest differential equation is $\frac{dy}{dx} = f(x)$, solved by a single antiderivative. Everything else in the course is a more elaborate version of “recover the function from information about its rate of change.”
- Separable equations: The separation method ends with an integral on each side. The integrals that appear are exactly the power, logarithm, exponential, and substitution forms practiced here.
- The role of $+ C$: The arbitrary constant becomes the place where an initial condition enters. A differential equation plus a starting value is called an initial-value problem, and solving it is integrate-then-determine-the-constant.
Toward Later Coursework
- First-order linear equations: Solving these multiplies through by an integrating factor and then integrates both sides, so fluency with integration is the rate-limiting skill.
- Modeling: A physical law states a rate of change. Translating that law into a function always passes through integration.
Audience Notes
For students who find math intimidating: You already know how to differentiate. Antidifferentiation is the same rules read in reverse. There is one new habit to build: always write $+ C$, and always check your answer by differentiating it. If differentiating returns the original, you are correct, with no guesswork.
For students interested in proof: The claim that any two antiderivatives differ by a constant is a consequence of the Mean Value Theorem. If $F' = G'$ on an interval, then $(F - G)' = 0$, and a function with zero derivative on an interval is constant there.
For students interested in careers: Antidifferentiation is how a measured rate becomes a total. Given a flow rate, integration gives accumulated volume; given an acceleration, integration gives velocity and then position. Every simulation that steps a system forward from rates is doing numerical antidifferentiation.
For gifted and curious students: Some perfectly ordinary functions have no antiderivative expressible with elementary formulas. The integral $\int e^{-x^2}\, dx$ is the famous example; it defines a new function (related to the error function) rather than reducing to known ones. This is why differential equations turns to geometric and numerical methods, not only symbolic ones.