{"id":4093,"date":"2026-07-14T12:48:08","date_gmt":"2026-07-14T12:48:08","guid":{"rendered":"https:\/\/www.mathros.net.ua\/en\/?p=4093"},"modified":"2026-07-14T15:43:11","modified_gmt":"2026-07-14T15:43:11","slug":"indefinite-integral","status":"publish","type":"post","link":"https:\/\/www.mathros.net.ua\/en\/indefinite-integral.html","title":{"rendered":"Indefinite Integral: Formula, Properties, and Examples"},"content":{"rendered":"

The indefinite integral is one of the fundamental topics in calculus<\/a>. It is closely related to the derivative<\/a>, but it works in the opposite direction. While a derivative shows how a function changes, an indefinite integral helps us recover a function from its derivative.<\/p>\n

At first, this topic may seem purely theoretical. However, it becomes much easier to understand once you recognize the main idea: we are not looking for a numerical value but for the general form of an antiderivative. This is why an arbitrary constant always appears after integration.<\/p>\n

Indefinite Integral: Definition and Basic Formula<\/h2>\n

Let us begin with the main definition. The indefinite integral of a function \\( f(x) \\) is the family of all antiderivatives of that function. In other words, we need to find a function \\( F(x) \\) whose derivative is equal to \\( f(x) \\).<\/p>\n

How do we write this? If \\( F'(x)=f(x) \\), then<\/p>\n

\\[
\n\\int f(x)\\,dx=F(x)+C.
\n\\]<\/p>\n

Here, the symbol \\( \\int \\) represents integration, the function \\( f(x) \\) is called the integrand, and ( dx ) shows the variable with respect to which the integration is performed. In addition, \\( C \\) is an arbitrary constant.<\/p>\n

Why is this constant necessary? The reason is that different functions can have the same derivative. For example, the functions \\( x^2 \\), \\( x^2+5 \\), and \\( x^2-10 \\) all have the same derivative, \\( 2\\cdot x \\). Therefore, when we find the integral of \\( 2\\cdot x \\), we cannot give only one answer. We need to write the general form of the antiderivative:<\/p>\n

\\[
\n\\int 2\\cdot x\\,dx=x^2+C.
\n\\]<\/p>\n

Thus, an indefinite integral does not represent one specific function. It describes all functions whose derivatives are equal to the given integrand.<\/p>\n

Therefore, integration can be viewed as the reverse operation of differentiation. But is it always enough simply to recognize an antiderivative? Of course not. More complex functions require special integration methods. However, the main principle remains the same: we need to find a function whose derivative gives the integrand.<\/p>\n

Properties of the Indefinite Integral: Explained Using Derivatives<\/h2>\n

Next, let us move on to the properties of the indefinite integral. They help simplify integrals and work with more complex integrands. Without these properties, even basic calculations would quickly become cumbersome.<\/p>\n

First, consider the constant multiple rule. If \\( k \\) is a constant, then<\/p>\n

\\[
\n\\int k\\cdot f(x)\\,dx=k\\cdot \\int f(x)\\,dx.
\n\\]<\/p>\n

Why is this true? Let \\( F(x) \\) be an antiderivative of \\( f(x) \\). Then<\/p>\n

\\[
\nF'(x)=f(x).
\n\\]<\/p>\n

Now consider the function \\( k\\cdot F(x) \\). Its derivative is<\/p>\n

\\[
\n(k\\cdot F(x))’=k\\cdot F'(x)=k\\cdot f(x).
\n\\]<\/p>\n

Therefore, \\( k\\cdot F(x) \\) is an antiderivative of \\( k\\cdot f(x) \\). This means that the indefinite integral of \\( k\\cdot f(x) \\) has the form<\/p>\n

\\[
\n\\int k\\cdot f(x)\\,dx=k\\cdot F(x)+C.
\n\\]<\/p>\n

Now let us look at the right-hand side of the rule. Since \\( F(x) \\) is an antiderivative of \\( f(x) \\), multiplying this antiderivative by the constant \\( k \\) gives \\( k\\cdot F(x) \\). We have just shown that \\( k\\cdot F(x) \\) is an antiderivative of \\( k\\cdot f(x) \\). Therefore, a constant factor can be taken outside the integral sign:<\/p>\n

\\[
\n\\int k\\cdot f(x)\\,dx=k\\cdot \\int f(x)\\,dx.
\n\\]<\/p>\n

Now consider the sum rule. The integral of the sum of two functions is equal to the sum of their integrals:<\/p>\n

\\[
\n\\int (f(x)+g(x))\\,dx=\\int f(x)\\,dx+\\int g(x)\\,dx.
\n\\]<\/p>\n

Let us prove this using derivatives. Let \\( F(x) \\) be an antiderivative of \\( f(x) \\), and let \\( G(x) \\) be an antiderivative of \\( g(x) \\). Then<\/p>\n

\\[
\nF'(x)=f(x), \\qquad G'(x)=g(x).
\n\\]<\/p>\n

Now consider the function \\( F(x)+G(x) \\). Its derivative is<\/p>\n

\\[
\n(F(x)+G(x))’=F'(x)+G'(x)=f(x)+g(x).
\n\\]<\/p>\n

Therefore, \\( F(x)+G(x) \\) is an antiderivative of \\( f(x)+g(x) \\). Thus,<\/p>\n

\\[
\n\\int (f(x)+g(x))\\,dx=F(x)+G(x)+C.
\n\\]<\/p>\n

On the other hand,<\/p>\n

\\[
\n\\int f(x)\\,dx=F(x)+C_1,
\n\\]<\/p>\n

and<\/p>\n

\\[
\n\\int g(x)\\,dx=G(x)+C_2.
\n\\]<\/p>\n

The sum of these integrals is then<\/p>\n

\\[
\nF(x)+G(x)+C_1+C_2.
\n\\]<\/p>\n

Since \\( C_1+C_2 \\) is also an arbitrary constant, we can denote it by \\( C \\). Therefore, we obtain the same formula:<\/p>\n

\\[
\n\\int (f(x)+g(x))\\,dx=\\int f(x)\\,dx+\\int g(x)\\,dx.
\n\\]<\/p>\n

The difference rule is proved in the same way:<\/p>\n

\\[
\n\\int (f(x)-g(x))\\,dx=\\int f(x)\\,dx-\\int g(x)\\,dx.
\n\\]<\/p>\n

Indeed, if \\( F'(x)=f(x) \\) and \\( G'(x)=g(x) \\), then<\/p>\n

\\[
\n(F(x)-G(x))’=F'(x)-G'(x)=f(x)-g(x).
\n\\]<\/p>\n

Therefore, \\( F(x)-G(x) \\) is an antiderivative of \\( f(x)-g(x) \\). This means that the integral of a difference can be written as the difference of the integrals.<\/p>\n

Thus, the main properties of the indefinite integral follow directly from the rules of differentiation. This is an important point. We are not simply memorizing formulas\u2014we can also see why they work. In the practical section, these properties will help us break more complex integrals into simpler and more convenient parts.<\/p>\n

Indefinite Integral in Practice: How to Apply Its Properties<\/h2>\n

Now let us move on to practice. Here, it is important not only to write down the answer but also to see exactly how the properties of the indefinite integral work. First, we will break the integrand into simpler parts and then integrate each part separately.<\/p>\n

Before solving the examples, let us recall the basic formula for integrating a power function:
\n\\[
\n\\int x^n\\,dx=\\frac{x^{n+1}}{n+1}+C,\\qquad n\\ne -1.
\n\\]
\nWe will also need the basic antiderivatives of trigonometric functions:
\n\\[
\n\\int \\cos(x)\\,dx=\\sin(x)+C,\\qquad \\int \\sin(x)\\,dx=-\\cos(x)+C.
\n\\]<\/p><\/blockquote>\n

Example 1. Find the indefinite integral
\n\\[
\n\\int 4\\cdot x^3\\,dx.
\n\\]<\/h3>\n

The function \\( x^3 \\) is multiplied by the constant \\( 4 \\). According to the constant multiple rule, we can take it outside the integral sign:<\/p>\n

\\[
\n\\int 4\\cdot x^3\\,dx=4\\cdot \\int x^3\\,dx.
\n\\]<\/p>\n

Now integrate the power function:<\/p>\n

\\[
\n\\int x^3\\,dx=\\frac{x^4}{4}.
\n\\]<\/p>\n

Therefore,<\/p>\n

\\[
\n4\\cdot \\int x^3\\,dx=4\\cdot \\frac{x^4}{4}+C.
\n\\]<\/p>\n

Simplify the constant factor:<\/p>\n

\\[
\n4\\cdot \\frac{x^4}{4}=x^4.
\n\\]<\/p>\n

Thus,<\/p>\n

\\[
\n\\int 4\\cdot x^3\\,dx=x^4+C.
\n\\]<\/p>\n

This example can easily be checked by differentiation:
\n\\[
\n(x^4+C)’=4\\cdot x^3.
\n\\]
\nThis confirms that the antiderivative we found corresponds to the integrand.<\/p><\/blockquote>\n

Example 2. Find the indefinite integral
\n\\[
\n\\int (3\\cdot x^2+2\\cdot x)\\,dx.
\n\\]<\/h3>\n

The integrand is the sum of two terms. Therefore, we use the sum rule:<\/p>\n

\\[
\n\\int (3\\cdot x^2+2\\cdot x)\\,dx=\\int 3\\cdot x^2\\,dx+\\int 2\\cdot x\\,dx.
\n\\]<\/p>\n

Now take the constant factor outside each integral:<\/p>\n

\\[
\n\\int 3\\cdot x^2\\,dx+\\int 2\\cdot x\\,dx
\n=
\n3\\cdot \\int x^2\\,dx+2\\cdot \\int x\\,dx.
\n\\]<\/p>\n

Find each antiderivative separately:<\/p>\n

\\[
\n\\int x^2\\,dx=\\frac{x^3}{3},
\n\\]<\/p>\n

and<\/p>\n

\\[
\n\\int x\\,dx=\\frac{x^2}{2}.
\n\\]<\/p>\n

Substitute these expressions:<\/p>\n

\\[
\n3\\cdot \\frac{x^3}{3}+2\\cdot \\frac{x^2}{2}+C.
\n\\]<\/p>\n

After simplifying, we get<\/p>\n

\\[
\nx^3+x^2+C.
\n\\]<\/p>\n

Thus,<\/p>\n

\\[
\n\\int (3\\cdot x^2+2\\cdot x)\\,dx=x^3+x^2+C.
\n\\]<\/p>\n

Example 3. Find the indefinite integral
\n\\[
\n\\int (5\\cdot x^4-6\\cdot x^2)\\,dx.
\n\\]<\/h3>\n

Here, we have the difference of two functions. Therefore, we first split the integral into two integrals:<\/p>\n

\\[
\n\\int (5\\cdot x^4-6\\cdot x^2)\\,dx
\n=
\n\\int 5\\cdot x^4\\,dx-\\int 6\\cdot x^2\\,dx.
\n\\]<\/p>\n

Next, take the constant factors outside the integral signs:<\/p>\n

\\[
\n\\int 5\\cdot x^4\\,dx-\\int 6\\cdot x^2\\,dx
\n=
\n5\\cdot \\int x^4\\,dx-6\\cdot \\int x^2\\,dx.
\n\\]<\/p>\n

Now find the antiderivatives of the power functions:<\/p>\n

\\[
\n\\int x^4\\,dx=\\frac{x^5}{5},\\qquad \\int x^2\\,dx=\\frac{x^3}{3}.
\n\\]<\/p>\n

Substitute them:<\/p>\n

\\[
\n5\\cdot \\frac{x^5}{5}-6\\cdot \\frac{x^3}{3}+C.
\n\\]<\/p>\n

Simplify each term:<\/p>\n

\\[
\n5\\cdot \\frac{x^5}{5}=x^5,\\qquad 6\\cdot \\frac{x^3}{3}=2\\cdot x^3.
\n\\]<\/p>\n

Therefore,<\/p>\n

\\[
\n\\int (5\\cdot x^4-6\\cdot x^2)\\,dx=x^5-2\\cdot x^3+C.
\n\\]<\/p>\n

Example 4. Find the indefinite integral
\n\\[
\n\\int (2\\cdot \\cos(x)+3\\cdot \\sin(x))\\,dx.
\n\\]<\/h3>\n

In this example, the integrand is the sum of trigonometric functions. Therefore, we first apply the sum rule:<\/p>\n

\\[
\n\\int (2\\cdot \\cos(x)+3\\cdot \\sin(x))\\,dx
\n=
\n\\int 2\\cdot \\cos(x)\\,dx+\\int 3\\cdot \\sin(x)\\,dx.
\n\\]<\/p>\n

Now take the constant factors outside the integrals:<\/p>\n

\\[
\n\\int 2\\cdot \\cos(x)\\,dx+\\int 3\\cdot \\sin(x)\\,dx
\n=
\n2\\cdot \\int \\cos(x)\\,dx+3\\cdot \\int \\sin(x)\\,dx.
\n\\]<\/p>\n

Recall the basic antiderivatives:<\/p>\n

\\[
\n\\int \\cos(x)\\,dx=\\sin(x),\\qquad \\int \\sin(x)\\,dx=-\\cos(x).
\n\\]<\/p>\n

Substitute these expressions:<\/p>\n

\\[
\n2\\cdot \\sin(x)+3\\cdot (-\\cos(x))+C.
\n\\]<\/p>\n

Thus,<\/p>\n

\\[
\n\\int (2\\cdot \\cos(x)+3\\cdot \\sin(x))\\,dx
\n=
\n2\\cdot \\sin(x)-3\\cdot \\cos(x)+C.
\n\\]<\/p>\n

Example 5. Find the indefinite integral
\n\\[
\n\\int (x^3-4\\cdot x+7)\\,dx.
\n\\]<\/h3>\n

The integrand consists of three terms. Therefore, we use the sum and difference rules:<\/p>\n

\\[
\n\\int (x^3-4\\cdot x+7)\\,dx
\n=
\n\\int x^3\\,dx-\\int 4\\cdot x\\,dx+\\int 7\\,dx.
\n\\]<\/p>\n

In the second integral, take the constant factor \\( 4 \\) outside:<\/p>\n

\\[
\n\\int x^3\\,dx-\\int 4\\cdot x\\,dx+\\int 7\\,dx
\n=
\n\\int x^3\\,dx-4\\cdot \\int x\\,dx+\\int 7\\,dx.
\n\\]<\/p>\n

Now find each antiderivative separately:<\/p>\n

\\[
\n\\int x^3\\,dx=\\frac{x^4}{4},\\qquad \\int x\\,dx=\\frac{x^2}{2},\\qquad \\int 7\\,dx=7\\cdot x.
\n\\]<\/p>\n

Substitute them:<\/p>\n

\\[
\n\\frac{x^4}{4}-4\\cdot \\frac{x^2}{2}+7\\cdot x+C.
\n\\]<\/p>\n

Simplify the middle term:<\/p>\n

\\[
\n4\\cdot \\frac{x^2}{2}=2\\cdot x^2.
\n\\]<\/p>\n

Thus,<\/p>\n

\\[
\n\\int (x^3-4\\cdot x+7)\\,dx
\n=
\n\\frac{x^4}{4}-2\\cdot x^2+7\\cdot x+C.
\n\\]<\/p>\n

In each example, we followed the same sequence of steps. First, we applied the properties of the indefinite integral. Then, we found the antiderivatives of simpler functions. Finally, we added the arbitrary constant \\( C \\). Following this sequence helps avoid confusion during calculations.<\/p><\/blockquote>\n

Further Steps in Integration: Topics to Explore<\/h2>\n

After learning about the indefinite integral, it is best to move forward gradually. The following topics will help you work with different types of integrals and understand which method to choose for a particular problem.<\/p>\n

    \n
  1. Table of Integrals: Basic Integration Formulas<\/a> \u2014 This article will cover the basic integrals most commonly used when solving problems.<\/li>\n
  2. Integration by Substitution: How the Method Works<\/a> \u2014 This article will explain how achange of variable helps simplify the integrand.<\/li>\n
  3. Integration by Parts: Formula and Step-by-Step Examples<\/a> \u2014 This article will examine a method that is especially useful for products of functions.<\/li>\n<\/ol>\n

    Indefinite Integral in Code: Turn the Algorithm into a Program<\/h2>\n

    If you enjoy programming, try using your knowledge of the indefinite integral not only in a notebook but also in your own code. The flowchart below shows an algorithm that accepts the coefficients of a polynomial, finds an antiderivative for each term, and produces the final answer as a complete expression.<\/p>\n

    Implement this algorithm in Pascal<\/em>, Python<\/em><\/a>, C++<\/em>, JavaScript<\/em>, or any other programming language you would like to work with. This task is a great way to combine the mathematical rule of integration with loops, conditions, arrays, and string processing.<\/p>\n

    \"Flowchart<\/p>\n","protected":false},"excerpt":{"rendered":"

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