What is an indefinite integral?

An indefinite integral is the reverse operation of differentiation. It represents the entire family of antiderivatives F(x)+C of a function f(x). The constant of integration C is an arbitrary constant, and the notation collectively expresses all functions sharing the same derivative.

Why is the constant of integration C necessary?

Functions that share the same derivative differ only by a constant. Without C, only one of the infinitely many antiderivatives is represented. Since an indefinite integral represents the entire family of antiderivatives, the constant of integration C makes this ambiguity explicit.

How can I remember the basic integral formulas?

Read the differentiation formulas in reverse. For example, reading (x^n)' = n x^{n-1} backwards gives the integral of x^n dx = x^{n+1}/(n+1)+C. The same approach works for trigonometric and exponential functions — each integral formula is derived from its corresponding differentiation formula.

Chapter 1: Indefinite Integral

Indefinite Integral

We study integration as the reverse operation of differentiation, and master the concept of antiderivatives along with the basic formulas.

1.1 Definition of Antiderivative

Definition: Antiderivative

For a function $f(x)$, a function $F(x)$ satisfying $F'(x) = f(x)$ is called an antiderivative (primitive function) of $f(x)$.

Differentiation and integration are inverse operations. If differentiating $F(x)$ yields $f(x)$, then integrating $f(x)$ recovers $F(x)$.

$f(x)$ $F(x)$ Differentiation $d/dx$ Integration $\int dx$

Fig. 1: Relationship between differentiation and integration

Theorem: Indeterminacy of Antiderivatives

If $F(x)$ is an antiderivative of $f(x)$, then $F(x) + C$ is also an antiderivative of $f(x)$ for any constant $C$.

Proof

Since $F'(x) = f(x)$, we have

$$\frac{d}{dx}(F(x) + C) = F'(x) + 0 = f(x)$$

Therefore $F(x) + C$ is also an antiderivative of $f(x)$. $\square$

Example 1.1

Find an antiderivative of $f(x) = 2x$.

Since $(x^2)' = 2x$, $F(x) = x^2$ is an antiderivative of $f(x) = 2x$.

Moreover, $x^2 + 1$, $x^2 - 5$, $x^2 + \pi$, etc., are all antiderivatives of $2x$.

1.2 Definition of Indefinite Integral

Definition: Indefinite Integral

The collection of all antiderivatives of $f(x)$ is called the indefinite integral of $f(x)$, written as:

$$\int f(x) \, dx = F(x) + C$$

Meaning of the notation:

$\int$: integral sign
$f(x)$: integrand
$dx$: indicates the variable of integration
$F(x)$: one antiderivative (primitive function) of $f(x)$
$C$: constant of integration — an arbitrary constant

Note: The result of an indefinite integral must always include the constant of integration $C$. Omitting $C$ means only one particular antiderivative is represented.

$y = F(x)$ $y = F(x)+1$ $y = F(x)+2$

Fig. 2: Geometric meaning of indefinite integral — family of antiderivatives

1.3 Table of Basic Formulas

By reading differentiation formulas in reverse, we obtain integration formulas. For detailed derivations of each formula, see the proof collection.

Power Functions

Integrand	Indefinite Integral	Condition	Derivation
$x^n$	$\dfrac{x^{n+1}}{n+1} + C$	$n \neq -1$（$n \notin \mathbb{Z}$ requires $x > 0$）	Proof
$\dfrac{1}{x}$	$\ln\|x\| + C$	$x \neq 0$	Proof

Exponential and Logarithmic Functions

Integrand	Indefinite Integral	Condition	Derivation
$e^x$	$e^x + C$		Proof
$a^x$	$\dfrac{a^x}{\ln a} + C$	$a > 0,\; a \neq 1$	Proof

Trigonometric Functions

Integrand	Indefinite Integral	Derivation
$\sin x$	$-\cos x + C$	Proof
$\cos x$	$\sin x + C$	Proof
$\sec^2 x = \dfrac{1}{\cos^2 x}$	$\tan x + C$	Proof
$\csc^2 x = \dfrac{1}{\sin^2 x}$	$-\cot x + C$	Proof

Integrals Involving Inverse Trigonometric Functions

Integrand	Indefinite Integral	Condition	Derivation
$\dfrac{1}{\sqrt{1-x^2}}$	$\arcsin x + C$	$\|x\| < 1$	Proof
$\dfrac{-1}{\sqrt{1-x^2}}$	$\arccos x + C$	$\|x\| < 1$	Proof
$\dfrac{1}{1+x^2}$	$\arctan x + C$		Proof
$\dfrac{1}{\|x\|\sqrt{x^2-1}}$	$\operatorname{arcsec} x + C$	$\|x\| > 1$	Proof

Hyperbolic Functions

Integrand	Indefinite Integral	Derivation
$\sinh x$	$\cosh x + C$	Proof
$\cosh x$	$\sinh x + C$	Proof
$\operatorname{sech}^2 x$	$\tanh x + C$	Proof
$\operatorname{csch}^2 x$	$-\coth x + C$	Proof

Integrals Involving Inverse Hyperbolic Functions

Integrand	Indefinite Integral	Condition	Derivation
$\dfrac{1}{\sqrt{x^2+1}}$	$\operatorname{arcsinh} x + C = \ln\!\bigl(x + \sqrt{x^2+1}\bigr) + C$		Proof
$\dfrac{1}{\sqrt{x^2-1}}$	$\operatorname{arccosh} x + C = \ln\!\bigl(x + \sqrt{x^2-1}\bigr) + C$	$x > 1$	Proof
$\dfrac{1}{1-x^2}$	$\operatorname{arctanh} x + C = \dfrac{1}{2}\ln\!\left\|\dfrac{1+x}{1-x}\right\| + C$	$\|x\| \neq 1$	Proof

1.4 Linearity

Theorem: Linearity of the Indefinite Integral

If $f(x)$ and $g(x)$ are integrable and $a$, $b$ are constants, then:

$$\int \{af(x) + bg(x)\} \, dx = a\int f(x) \, dx + b\int g(x) \, dx$$

Proof

Let $F(x)$ be an antiderivative of $f(x)$ (i.e., $F'(x) = f(x)$) and $G(x)$ be an antiderivative of $g(x)$ ($G'(x) = g(x)$).

Then:

$$\frac{d}{dx}\{aF(x) + bG(x)\} = aF'(x) + bG'(x) = af(x) + bg(x)$$

Therefore $aF(x) + bG(x)$ is an antiderivative of $af(x) + bg(x)$. $\square$

In particular:

$\displaystyle\int kf(x) \, dx = k\int f(x) \, dx$ (scalar multiple)
$\displaystyle\int \{f(x) + g(x)\} \, dx = \int f(x) \, dx + \int g(x) \, dx$ (integral of a sum)

Example 1.2

Evaluate $\displaystyle\int (3x^2 + 2x + 1) \, dx$.

$$\begin{align} \int (3x^2 + 2x + 1) \, dx &= 3\int x^2 \, dx + 2\int x \, dx + \int 1 \, dx \\ &= 3 \cdot \frac{x^3}{3} + 2 \cdot \frac{x^2}{2} + x + C \\ &= x^3 + x^2 + x + C \end{align}$$

1.5 Worked Examples

Example 1.3: Power Function Integration

Evaluate $\displaystyle\int x^5 \, dx$.

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

Verification: $\left(\dfrac{x^6}{6}\right)' = \dfrac{6x^5}{6} = x^5$ ✓

Example 1.4: Fractional Power Integration

Evaluate $\displaystyle\int \sqrt{x} \, dx$.

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

$$\int x^{1/2} \, dx = \frac{x^{1/2+1}}{1/2+1} + C = \frac{x^{3/2}}{3/2} + C = \frac{2}{3}x^{3/2} + C$$

Alternative form: $\dfrac{2}{3}x\sqrt{x} + C$

Example 1.5: Negative Power Integration

Evaluate $\displaystyle\int \frac{1}{x^3} \, dx$.

Rewriting $\dfrac{1}{x^3} = x^{-3}$:

$$\int x^{-3} \, dx = \frac{x^{-3+1}}{-3+1} + C = \frac{x^{-2}}{-2} + C = -\frac{1}{2x^2} + C$$

Example 1.6: Combination of Trigonometric Functions

Evaluate $\displaystyle\int (2\sin x - 3\cos x) \, dx$.

$$\begin{align} \int (2\sin x - 3\cos x) \, dx &= 2\int \sin x \, dx - 3\int \cos x \, dx \\ &= 2(-\cos x) - 3(\sin x) + C \\ &= -2\cos x - 3\sin x + C \end{align}$$

Example 1.7: Expanding a Polynomial Before Integration

Evaluate $\displaystyle\int (x+1)^2 \, dx$.

Expand first, then integrate:

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

Note: Using the substitution method (covered later), one can also obtain $\dfrac{(x+1)^3}{3} + C$ without expanding.

1.6 Exercises

Problem 1

Evaluate the following indefinite integrals.

$\displaystyle\int x^4 \, dx$
$\displaystyle\int \frac{1}{x^2} \, dx$
$\displaystyle\int \sqrt[3]{x} \, dx$
$\displaystyle\int 3e^x \, dx$

Show Solution

$\displaystyle\int x^4 \, dx = \frac{x^5}{5} + C$
$\displaystyle\int \frac{1}{x^2} \, dx = \int x^{-2} \, dx = \frac{x^{-1}}{-1} + C = -\frac{1}{x} + C$
$\displaystyle\int \sqrt[3]{x} \, dx = \int x^{1/3} \, dx = \frac{x^{4/3}}{4/3} + C = \frac{3}{4}x^{4/3} + C$
$\displaystyle\int 3e^x \, dx = 3\int e^x \, dx = 3e^x + C$

Problem 2

Evaluate the following indefinite integrals.

$\displaystyle\int (x^3 - 4x + 2) \, dx$
$\displaystyle\int \left(x + \frac{1}{x}\right)^2 \, dx$
$\displaystyle\int (2\sin x + \cos x) \, dx$

Show Solution

$\displaystyle\int (x^3 - 4x + 2) \, dx = \frac{x^4}{4} - 2x^2 + 2x + C$
First expand: $\left(x + \dfrac{1}{x}\right)^2 = x^2 + 2 + \dfrac{1}{x^2}$

$\displaystyle\int \left(x^2 + 2 + x^{-2}\right) \, dx = \frac{x^3}{3} + 2x - \frac{1}{x} + C$
$\displaystyle\int (2\sin x + \cos x) \, dx = -2\cos x + \sin x + C$

Problem 3

Find the function $F(x)$ satisfying $F'(x) = 4x^3 - 6x^2 + 2$ and $F(0) = 3$.

Show Solution

First compute the indefinite integral:

$$F(x) = \int (4x^3 - 6x^2 + 2) \, dx = x^4 - 2x^3 + 2x + C$$

From the initial condition $F(0) = 3$:

$$F(0) = 0 - 0 + 0 + C = C = 3$$

Therefore:

$$F(x) = x^4 - 2x^3 + 2x + 3$$

References

Rudin, W. (1976). Principles of Mathematical Analysis (3rd ed.). McGraw-Hill. Chapter 6.
Stewart, J. (2015). Calculus: Early Transcendentals (8th ed.). Cengage Learning. Chapter 4.
Wikipedia. "Antiderivative".