What Are Indefinite Integrals?
Indefinite integrals are the reverse of differentiation: the indefinite integral of a function $f(x)$ is the collection of every function whose derivative is $f(x)$. It is written
$$\int f(x),dx = F(x) + C,$$
where $F$ is any antiderivative of $f$ (a function with $F'(x) = f(x)$) and $C$ is the constant of integration. The symbol $\int$ is the integral sign, $f(x)$ is the integrand, and $dx$ names the variable you are integrating with respect to.
The single most important word here is family. Differentiation destroys constants, because the derivative of any constant is zero. So if $F(x)$ works, then $F(x) + 1$, $F(x) - 7$, and $F(x) + C$ for every real $C$ all differentiate back to the same $f(x)$. The indefinite integral gathers all of them at once.
$$\frac{d}{dx}\big(x^2\big) = 2x, \qquad \frac{d}{dx}\big(x^2 + 5\big) = 2x, \qquad \text{so} \qquad \int 2x,dx = x^2 + C$$
Because the output is a whole function (strictly, a family of functions) rather than a single value, an indefinite integral is sometimes called the general antiderivative. Every one of the standard rules below is just a differentiation fact read backwards. For the collection of antiderivatives on their own terms, see the antiderivatives reference.
How Is An Indefinite Integral Different From A Definite Integral?
This is the distinction students lose the most marks on, so it is worth stating plainly before any computation.
An indefinite integral $\int f(x),dx$ has no limits. It evaluates to a function plus a constant, $F(x) + C$, and represents a family of curves.
A definite integral $\int_a^b f(x),dx$ has a lower limit $a$ and an upper limit $b$. It evaluates to a single number, the signed area under the curve, and carries no $+C$.
The two are linked by the fundamental theorem of calculus: you compute a definite integral by first finding an indefinite one (an antiderivative $F$) and then subtracting, $\int_a^b f(x),dx = F(b) - F(a)$. The constant cancels in that subtraction, which is exactly why the definite integral never keeps a $+C$.
Table: Indefinite versus definite integrals at a glance.
Feature | Indefinite integral | Definite integral |
|---|---|---|
Notation | $\int f(x),dx$ | $\int_a^b f(x),dx$ |
Result | A function $F(x) + C$ | A number |
Constant | Always carries $+C$ | No $+C$ (it cancels) |
Meaning | Family of antiderivatives | Signed area over $[a, b]$ |
Keep this contrast in mind, because treating one like the other is a ranked common mistake, covered near the end.
What Are The Basic Rules For Indefinite Integrals?
Every rule below is a derivative fact reversed, and each keeps a single $+C$ on the final answer no matter how many terms it has. These standard forms cover most first-year integrals; the full inventory lives in the basic integration formulas and the larger table of integrals.
Constant-multiple rule. A constant factor slides straight through the integral sign: $$\int k,f(x),dx = k\int f(x),dx$$
Sum and difference rule. An integral of a sum is the sum of the integrals: $$\int \big[f(x) \pm g(x)\big],dx = \int f(x),dx \pm \int g(x),dx$$
Power rule for integration. 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 $n = -1$ exception. The power rule would divide by zero when $n = -1$, so that one case uses the natural logarithm instead: $$\int \frac{1}{x},dx = \ln\lvert x\rvert + C$$
Exponential and trigonometric forms. Each of these is its matching derivative read backwards: $$\int e^x,dx = e^x + C, \quad \int \sin x,dx = -\cos x + C, \quad \int \cos x,dx = \sin x + C, \quad \int \sec^2 x,dx = \tan x + C$$
Table: The core indefinite-integral rules, each verified by differentiating the right side.
Integral | Result | Check (differentiate the result) |
|---|---|---|
$\int x^n,dx$ ($n \neq -1$) | $\dfrac{x^{n+1}}{n+1} + C$ | $\dfrac{(n+1)x^{n}}{n+1} = x^n$ |
$\int \dfrac{1}{x},dx$ | $\ln\lvert x\rvert + C$ | $\dfrac{1}{x}$ |
$\int e^x,dx$ | $e^x + C$ | $e^x$ |
$\int \sin x,dx$ | $-\cos x + C$ | $\sin x$ |
$\int \cos x,dx$ | $\sin x + C$ | $\cos x$ |
$\int \sec^2 x,dx$ | $\tan x + C$ | $\sec^2 x$ |
The right-hand column is the whole trick: an indefinite integral is correct exactly when differentiating it returns the integrand. That single test settles every worked example below.
How Do You Solve Indefinite Integrals? Worked Examples
Each example is fully stepped, and every answer is checked by differentiating it back to the original integrand.
Example 1: A polynomial.
Find $\displaystyle\int \big(3x^2 + 4x - 5\big),dx$.
Integrate term by term with the sum rule and the power rule, treating the constant $-5$ as $-5x^0$:
$$\int \big(3x^2 + 4x - 5\big),dx = \frac{3x^{3}}{3} + \frac{4x^{2}}{2} - 5x + C = x^3 + 2x^2 - 5x + C$$
Check: $\dfrac{d}{dx}\big(x^3 + 2x^2 - 5x + C\big) = 3x^2 + 4x - 5$, the original integrand.
Final answer: $x^3 + 2x^2 - 5x + C$.
Example 2: A constant multiple and the $1/x$ case together.
Find $\displaystyle\int \left(2\cos x + \frac{6}{x}\right),dx$.
Split with the sum rule, pull the constants out, and use $\int \cos x,dx = \sin x$ with the $n = -1$ form $\int \frac{1}{x},dx = \ln\lvert x\rvert$:
$$\int \left(2\cos x + \frac{6}{x}\right),dx = 2\sin x + 6\ln\lvert x\rvert + C$$
Check: $\dfrac{d}{dx}\big(2\sin x + 6\ln\lvert x\rvert + C\big) = 2\cos x + \dfrac{6}{x}$, which matches.
Final answer: $2\sin x + 6\ln\lvert x\rvert + C$.
Example 3: A root, rewritten as a power.
Find $\displaystyle\int \sqrt{x},dx$.
Rewrite the root as a fractional exponent, $\sqrt{x} = x^{1/2}$, then apply the power rule with $n = \tfrac{1}{2}$:
$$\int x^{1/2},dx = \frac{x^{3/2}}{3/2} + C = \frac{2}{3}x^{3/2} + C$$
Check: $\dfrac{d}{dx}\left(\dfrac{2}{3}x^{3/2} + C\right) = \dfrac{2}{3}\cdot\dfrac{3}{2}x^{1/2} = x^{1/2} = \sqrt{x}$.
Final answer: $\dfrac{2}{3}x^{3/2} + C$.
What Methods Solve Harder Indefinite Integrals?
The rules above handle sums of powers, exponentials, and the basic trigonometric functions directly. When the integrand is a composition or a product, you first rewrite it into a form the rules can reach. Two techniques do most of that work, and each reverses a differentiation rule.
Integration by substitution reverses the chain rule. You replace an inner function with a new variable $u$ so the integral collapses to a standard form, then substitute back. It is the go-to move for integrals such as $\int 2x,e^{x^2},dx$. See integration by substitution.
Integration by parts reverses the product rule. It rewrites $\int u,dv$ as $uv - \int v,du$, trading a hard product for an easier integral, and handles cases such as $\int x\cos x,dx$. See integration by parts.
A wider survey of when to reach for each technique, including partial fractions and trigonometric substitution, sits in methods of integration. All of them lean on the same idea that integration and differentiation and integration are inverse operations, so every method is a derivative rule run in reverse.
Not every integral has an answer in elementary functions. The integrals $\int e^{-x^2},dx$ and $\int \frac{\sin x}{x},dx$ are perfectly well defined, yet no combination of powers, roots, exponentials, logarithms, and trigonometric functions differentiates back to them. Their antiderivatives exist but are given names of their own rather than a closed formula, so writing a fake elementary answer for either is simply wrong.
Why Do Indefinite Integrals Need A Constant?
The $+C$ is not decoration, and it is not optional. It is forced by a fact about derivatives.
Constants vanish under differentiation. Since $\frac{d}{dx}(C) = 0$, adding any constant to an antiderivative leaves its derivative unchanged. So a single formula could never capture "all functions with this derivative"; you need the constant to hold the whole family.
Any two antiderivatives differ by a constant. If $F' = f$ and $G' = f$ on an interval, then $(F - G)' = 0$, and a function with zero derivative everywhere on an interval is constant. So $F - G = C$, which means the family $F(x) + C$ genuinely contains every antiderivative, not just some.
The geometry is a stack of parallel curves. Graph $y = x^2$, $y = x^2 + 1$, and $y = x^2 - 2$. They are identical shapes shifted vertically, and at any fixed $x$ their tangents are parallel because they share the slope function $2x$. The indefinite integral names that entire stack in one line, and the $+C$ chooses which curve you land on.
Read together, the algebra (constants have zero derivative) and the geometry (curves that differ only in height) are the same statement. The constant of integration is what turns one antiderivative into all of them. For the deeper treatment of that single symbol, see the constant of integration.
Who Invented Indefinite Integrals?
The idea of reversing a rate to recover a total is old, but the notation and the theory that made it routine arrived in the late seventeenth century.
Two later figures made the subject rigorous:
Augustin-Louis Cauchy (1789–1857, France) gave integration a precise definition built on limits, so an antiderivative rested on exact reasoning rather than intuition about infinitely small pieces.
Bernhard Riemann (1826–1866, Germany) sharpened the definite integral into the limit of sums that now carry his name, pinning down exactly which functions can be integrated at all.
Where Are Indefinite Integrals Used In The Real World?
Any time you know a rate of change and want the quantity behind it, you are computing an indefinite integral. The $+C$ is the piece of information the rate alone cannot tell you, usually an initial value.
Physics and motion: velocity is the antiderivative of acceleration, and position is the antiderivative of velocity. The constant is the starting position or starting speed, fixed by where the object began.
Engineering: the charge stored on a capacitor is the antiderivative of the current flowing into it over time, with the constant set by the initial charge.
Economics: total cost is the antiderivative of marginal cost, so a firm reconstructs its cost function from the per-unit rate it can measure, with the fixed cost playing the role of the constant.
Biology and medicine: a population's size is the antiderivative of its growth rate, and the constant fixes the population you started counting from.
Earth science: the volume of water stored behind a dam is the antiderivative of the net inflow rate, with the constant set by the reservoir's level at the start of the measurement.
One move, reversing a rate, rebuilds distance, charge, cost, population, and volume alike. The initial condition that pins down the constant is what turns the general antiderivative into the specific answer a problem asks for.
What Are The Most Common Mistakes With Indefinite Integrals?
These three errors cause most lost marks on indefinite integrals, verified against Paul's Online Notes "Common Math Errors" and its section on constant-of-integration subtleties, the University of Wisconsin "Definite versus Indefinite Integral" review, and recurring r/calculus and r/learnmath questions.
Dropping the constant of integration.
Where it slips in:
A student computes $\int 2x,dx = x^2$ and stops, leaving off the $+C$ because the arithmetic looked finished.
Don't do this:
Do not write an indefinite integral without $+C$. The answer is a family of functions, and $x^2$ alone names only one member of it.
The correct way:
Attach $+C$ to every indefinite integral: $\int 2x,dx = x^2 + C$. One constant covers the whole expression, however many terms it has.
Using the power rule at $n = -1$.
Where it slips in:
A student integrates $\frac{1}{x} = x^{-1}$ with the power rule and writes $\dfrac{x^{0}}{0} + C$, which divides by zero.
Don't do this:
Do not apply $\int x^n,dx = \dfrac{x^{n+1}}{n+1} + C$ when $n = -1$. The formula is undefined there.
The correct way:
Use the logarithm form for that one case: $\int \dfrac{1}{x},dx = \ln\lvert x\rvert + C$. The absolute value keeps the answer valid for negative $x$ as well.
Treating an indefinite integral like a definite one.
Where it slips in:
A student "evaluates" $\int x^2,dx$ by plugging in numbers, or expects a single value, blurring it with $\int_a^b x^2,dx$.
Don't do this:
Do not force a number out of an indefinite integral, and do not carry a $+C$ into a definite one. An indefinite integral is a function; a definite integral is a number.
The correct way:
Return a function for the indefinite case, $\int x^2,dx = \dfrac{x^3}{3} + C$, and only substitute limits when the integral actually has them, where the constant cancels.
Practice Problems On Indefinite Integrals
Work each one with the rules above, then check by differentiating your answer back to the integrand. Answers are verified.
Find $\int \big(4x^3 - 2x\big),dx$.
(Answer: $x^4 - x^2 + C$.)Find $\int 5,dx$. (Answer: $5x + C$.)
Find $\int \dfrac{1}{x^2},dx$.
(Answer: rewrite as $\int x^{-2},dx = -\dfrac{1}{x} + C$.)Find $\int \big(e^x + \sin x\big),dx$.
(Answer: $e^x - \cos x + C$.)Find $\int 6\sqrt{x},dx$.
(Answer: $4x^{3/2} + C$.)Find $\int \dfrac{3}{x},dx$.
(Answer: $3\ln\lvert x\rvert + C$.)
Where Should You Go Next After Indefinite Integrals?
Indefinite integrals open straight into the rest of integral calculus, and a few natural doors lead onward.
Constant of integration. Go deeper on the $+C$, initial conditions, and how one extra fact pins down a specific antiderivative.
Definite integrals. Add limits to turn a family of functions into a single number, the signed area under a curve.
Integration by substitution. The first real technique for integrals the basic rules cannot reach directly.
Fundamental theorem of calculus. The result that ties indefinite and definite integrals together as two sides of one idea.
If your child is meeting indefinite integrals for the first time, a live Bhanzu trainer teaches them from the antiderivative picture up, so the $+C$ feels inevitable rather than a rule to memorise, in the Bhanzu math classes.
Was this article helpful?
Your feedback helps us write better content