Have you noticed how members of the same family tend to look like each other? The same is true for families of functions! Functions of the same form look very similar to each other, like members of the same family. Indefinite integrals are no different here. They are a family of antiderivatives of a function, so they look very similar to each other, too.
In this article, you will learn about what an indefinite integral is, its definition, formula, and properties. You'll also see examples of calculations of indefinite integrals.
Indefinite Integral Definition
As you know from the antiderivatives article, the process of finding a function's antiderivative is called integration. Remember that, if you are given a function, \( f(x) \), an antiderivative of \( f(x) \) is any function \( F(x) \) that satisfies the condition:
\[ F'(x) = f(x). \]
So, how does the indefinite integral come into play here?
Well, it is used to reference the entire family of antiderivatives of a function, whereas an antiderivative is just one of infinite possibilities.
With this in mind, you define the indefinite integral as:
If \( F(x) \) is an antiderivative of a function \( f(x) \), then the family of antiderivatives of \( f(x) \) is called an indefinite integral. The notation for this indefinite integral is:
\[ \int f(x) ~\mathrm{d}x = F(x) + C, \]
where \(C\) is any constant.
Note that:
\( \int \) is called the integral symbol,
\( f(x) \) is called the integrand,
\( x \) is called the integration variable,
\( \mathrm{d}x \) is called the differential,
\( F(x) \) is the antiderivative, and
\( C \) is called the integration constant (or constant of integration).
Note that the terms “indefinite integral” and “antiderivative” are sometimes used interchangeably, and, in some texts, an antiderivative is also called a “primitive function”.
Given the terminology introduced to you in this definition, the act of finding the antiderivatives of a function, \( f \), is commonly referred to as either:
- integrating \( \mathbf{f} \) or
- finding the integral of \( \mathbf{f} \).
For a function, \( f(x) \), and its antiderivative, \( F(x) \), the functions of the form \( F(x) + C \), where \( C \) is any constant, are often referred to as the family of antiderivatives of \( \mathbf{f(x)} \).
Indefinite Integral: A Family of Antiderivatives
To help visualize what “family of antiderivatives” means, consider this example.
The Indefinite Integral, the Integration Constant, and the Family of Antiderivatives
Consider the function:
\[ f(x) = 2x. \]
What is the indefinite integral of \( f(x) \)?
Solution:
The indefinite integral of \( f(x) = 2x \) is
\[ \int 2x ~\mathrm{d}x = x^{2} + C, \]
where \(C\) is the integration constant.
Since the derivative of any constant is \(0\), \(C\) can be any constant (as long as it is a real number), whether positive, negative, or even \(0\) itself.
When you find an indefinite integral, you always add the integration constant, \(C\), to your final solution.
Why?
Because you have infinite solutions – the family of antiderivatives of \( f(x) \).
- For this example, the collection of all functions of the form \( F(x) = x^{2} + C \) (where \(C\) is any constant) is known as the family of antiderivatives of the function \( f(x) = 2x \).
Put another way, an indefinite integral doesn’t have any limits, so you’re finding a set of integrals, instead of a specific one (as in the case of solving definite integrals). The \(+\,C\) indicates that the solution has infinite possibilities.
Just a few of the possible solutions for the indefinite integral of \( f(x) = 2x \) are shown in the graphs below.
Figure 1. The family of antiderivatives of the function \( f(x) = 2x \) consists of all functions of the form \( f(x) = x^{2}+C \), where \(C\) is any constant (as long as it is a real number).
Indefinite Integral Formula
Just as with antiderivatives in general, indefinite integrals do not have just one formula for solving them. There are a variety of rules and properties that you will learn to use to solve indefinite integrals – they are based on the differentiation rules you have already learned. The reason for this is discussed in the fundamental theorem of calculus article.
That being said, the essence of finding an indefinite integral of a function is to do the reverse of the differentiation rules you already know.
Properties of the Indefinite Integral
Since the indefinite integral is just a family of antiderivatives, their properties are the same. But, to reiterate, the indefinite integral is linear; i.e., you can integrate “term by term” for sums, differences, and constant multiples. These linearity properties are summarized by the rules below.
Sum/Difference Property:
\[ \int (f(x) \pm g(x)) ~\mathrm{d}x = \int f(x) ~\mathrm{d}x \pm \int g(x) ~\mathrm{d}x \]
Constant Multiple Property:
\[ \int kf(x) ~\mathrm{d}x = k \int f(x) ~\mathrm{d}x \]
Proofs of the Properties of the Indefinite Integral
- In general, if \( F \) is an antiderivative of \( f\) and \( G \) is an antiderivative of \( g \), then\[ \frac{d}{dx} (F(x) \pm G(x)) = F'(x) \pm G'(x) = f(x) \pm g(x). \]This means that \( F(x) \pm G(x) \) is an antiderivative of \( f(x) \pm g(x) \), so you have\[ \int (f(x) \pm g(x)) ~\mathrm{d}x = F(x) \pm G(x) + C. \]
- Now consider finding an antiderivative of \( kf(x) \), where \( k \) is any constant. Since you know that\[ \frac{d}{dx} (kf(x)) = k \frac{d}{dx}F(x) = kf'(x) \]for any constant \( k \), you can conclude that\[ \int kf(x) ~\mathrm{d}x = kF(x) + C. \]
Rules for Indefinite Integrals
For the most part, the rules for finding the indefinite integral of a function are the inverse (or reverse) of the rules for finding derivatives.
Below is a list of rules for common indefinite integrals.
The Constant RuleIf you consider the function \( F(x) = 3 \) and write its derivative as \( f(x) \), this means that \( f(x) = \frac{dF}{dx} \). You already know that you can find the derivative of this function by applying the constant rule for derivatives: \( \frac{d}{dx}(k) = 0 \).Now say you want to reverse this process and ask yourself, which function(s) could possibly have \( f(x) = 0 \) as a derivative? Obviously, \( F(x) = 3 \) is one answer. You say that \( F(x) = 3 \) is an antiderivative of \( f(x) = 0 \).
However, there are other functions whose derivative is \( f(x) = 0 \), including but not limited to \( F(x) = 5 \), \( F(x) = -4 \), and \( F(x) = 200 \). This is because when you take a derivative, the constant disappears.
Therefore, if you are given any antiderivative of \( f(x) \), all others can be found by adding a different constant. In other words, if \( F(x) \) is an antiderivative of \( f(x) \), then \( F(x) + C \) is also an antiderivative of \( f(x) \) for any constant \( C \). This group, or family, of antiderivatives is represented by the indefinite integral.
This thought process is what brings you to the indefinite integral rules.
\[ \begin{align}\text{Derivative Rule: } &\frac{d}{dx}(k) = 0 \\\text{Indefinite Integral Rule: } &\int k ~\mathrm{d}x = kx + C\end{align} \]
The Power RuleContinuing the thought process from above, you can see how most of these indefinite integral rules work. For many functions, evaluating the indefinite integral is the direct opposite of the derivative. For instance, if \( n \neq -1 \),\[ \frac{d}{dx} \left( \frac{x^{n+1}}{n+1} \right) = (n+1) \frac{x^{n}}{n+1} = x^{n}, \]leads you directly to the power rule for indefinite integrals. So, the derivative and indefinite integral rules are:
\[ \begin{align}\text{Derivative Rule: } &\frac{d}{dx}\left(x^{n}\right) = nx^{n-1} \\\text{Indefinite Integral Rule: } &\int x^{n} ~\mathrm{d}x = \frac{x^{n+1}}{n+1} + C, n \neq -1\end{align} \]
The Natural Logarithm Rule
\[ \begin{align}\text{Derivative Rule: } &\frac{d}{dx}(\ln|x|) = \frac{1}{x} \\\text{Indefinite Integral Rule: } &\int \frac{1}{x} ~\mathrm{d}x = \ln|x| + C\end{align} \]
The Exponential Rule (with base \( e \))
\[ \begin{align}\text{Derivative Rule: } &\frac{d}{dx}\left(e^{x}\right) = e^{x} \\\text{Indefinite Integral Rule: } &\int e^{x} ~\mathrm{d}x = e^{x} + C\end{align} \]
The Exponential Rule (with base \( a \))
\[ \begin{align}\text{Derivative Rule: } &\frac{d}{dx}\left(a^{x}\right) = a^{x} \ln a \\\text{Indefinite Integral Rule: } &\int a^{x} ~\mathrm{d}x = \frac{a^{x}}{\ln a} + C, ~\ a \neq 1\end{align} \]
The Sine Rule
\[ \begin{align}\text{Derivative Rule: } &\frac{d}{dx}(\sin(x)) = \cos(x) \\\text{Indefinite Integral Rule: } &\int \cos(x) ~\mathrm{d}x = \sin(x) + C\end{align} \]
The Cosine Rule
\[ \begin{align}\text{Derivative Rule: } &\frac{d}{dx}(\cos(x)) = -\sin(x) \\\text{Indefinite Integral Rule: } &\int \sin(x) ~\mathrm{d}x = -\cos(x) + C\end{align} \]
The Tangent Rule
\[ \begin{align}\text{Derivative Rule: } &\frac{d}{dx}(\tan(x)) = \sec^{2}(x) \\\text{Indefinite Integral Rule: } &\int \sec^{2}(x) ~\mathrm{d}x = \tan(x) + C\end{align} \]
The Cosecant Rule
\[ \begin{align}\text{Derivative Rule: } &\frac{d}{dx}(\csc(x)) = -\csc(x)\cot(x) \\\text{Indefinite Integral Rule: } &\int \csc(x)\cot(x) ~\mathrm{d}x = -\csc(x) + C\end{align} \]
The Secant Rule
\[ \begin{align}\text{Derivative Rule: } &\frac{d}{dx}(\sec(x)) = \sec(x)\tan(x) \\\text{Indefinite Integral Rule: } &\int \sec(x)\tan(x) ~\mathrm{d}x = \sec(x) + C\end{align} \]
The Cotangent Rule
\[ \begin{align}\text{Derivative Rule: } &\frac{d}{dx}(\cot(x)) = -\csc^{2}(x) \\\text{Indefinite Integral Rule: } &\int \csc^{2}(x) ~\mathrm{d}x = -\cot(x) + C\end{align} \]
The Inverse Sine Rule
\[ \begin{align}\text{Derivative Rule: } &\frac{d}{dx}\left(\sin^{-1}(x)\right) = \frac{1}{\sqrt{1-x^{2}}} \\\text{Indefinite Integral Rule: } &\int \frac{1}{\sqrt{1-x^{2}}} ~\mathrm{d}x = \sin^{-1}(x) + C\end{align} \]
The Inverse Tangent Rule
\[ \begin{align}\text{Derivative Rule: } &\frac{d}{dx}\left(\tan^{-1}(x)\right) = \frac{1}{1+x^{2}} \\\text{Indefinite Integral Rule: } &\int \frac{1}{1+x^{2}} ~\mathrm{d}x = \tan^{-1}(x) + C\end{align} \]
The Inverse Secant Rule
\[ \begin{align}\text{Derivative Rule: } &\frac{d}{dx}\left(\sec^{-1}(x)\right) = \frac{1}{x\sqrt{x^{2}-1}} \\\text{Indefinite Integral Rule: } &\int \frac{1}{x\sqrt{x^{2}-1}} ~\mathrm{d}x = \sec^{-1}|x| + C\end{align} \]
Note that in integral notation, you can treat the differential, \( \mathrm{d}x \), as a movable variable. This means you could, for example, rewrite the \( 3^{rd} \) rule from the list above as:
\[ \int \frac{1}{x} ~\mathrm{d}x = \ln|x| + C \Rightarrow \int \frac{\mathrm{d}x}{x} = \ln|x| + C \]
Indefinite Integrals: Mistakes to Avoid
Did you notice in the list above that there are no product, quotient, or chain rules for integrals?
What does this mean?
It means that, just like with derivatives, the rules that apply to addition and subtraction, do not apply in the same way to multiplication and division. In other words, just like with derivatives:
- The integral of the product (or quotient) of two functions is not equal to the product (or quotient) of the integral of the functions.\[ \begin{align}\int f(x) \cdot g(x) ~\mathrm{d}x &\neq \int f(x) ~\mathrm{d}x \cdot \int g(x) ~\mathrm{d}x \\\int \frac{f(x)}{g(x)} ~\mathrm{d}x &\neq \frac{\int f(x) ~\mathrm{d}x}{\int g(x) ~\mathrm{d}x}\end{align} \]
Instead:
the product and quotient rules for derivatives lead you to integration by parts, and
the chain rule for derivatives leads you to integration by substitution.
While integration by parts is derived specifically from the product rule for derivatives, it applies to both a product and a quotient of integrals. This is because, for any two functions \( f \) and \( g \), you can write the quotient of the two functions as a product:
\[ \frac{f}{g} = f \cdot \frac{1}{g}. \]
In other words, you can think of the quotient rule for derivatives as a product rule in disguise; the same holds true for integration by parts.
For example:
Consider the indefinite integral
\[ \int \frac{\sin\left(\frac{1}{x}\right)}{x^{2}} ~\mathrm{d}x. \]
You solve this integral using integration by parts. However, instead of using a quotient rule, it is easier to rewrite this integral as
\[ \int \sin\left(\frac{1}{x}\right) \cdot \frac{1}{x^{2}} ~\mathrm{d}x \]
and use a product rule to perform the integration by parts.
Indefinite Integral Calculation
When it comes to calculating an indefinite integral, the exact steps you take will depend on the integral itself. However, there are some very basic steps that you will need to remember for calculating all indefinite integrals.
Basic Steps to Calculate an Indefinite Integral
Determine which properties and rules apply.
If you need to use more than one property or rule, decide the order in which to use them.
Use the rules you decided on.
Add the integration constant.
Verify your result by proving that \( F'(x) = f(x) \).
Indefinite Integral Examples
In the following examples, evaluate each of the indefinite integrals. This first example is relatively simple.
Evaluate
\[ \int \left( 4x^{3} - 6x^{2} + 2x + 5 \right) ~\mathrm{d}x \]
Solution:
Determine which properties and rules apply.
If you need to use more than one property or rule, decide the order in which to use them.
Apply the sum/difference rule for integrals.
Apply the constant multiple rule for integrals.
Apply the power rule for integrals.
Use the rules you decided on.
Apply the sum/difference rule for integrals by rewriting the integral as\[ \begin{align}\int &\left( 4x^{3} - 6x^{2} + 2x + 5 \right) ~\mathrm{d}x = \\&\int 4x^{3} ~\mathrm{d}x - \int 6x^{2} ~\mathrm{d}x + \int 2x ~\mathrm{d}x + \int 5 ~\mathrm{d}x.\end{align} \]
Apply the constant multiple rule for integrals by rewriting the integral as\[ \begin{align}\int &\left( 4x^{3} - 6x^{2} + 2x + 5 \right) ~\mathrm{d}x = \\&4 \int x^{3} ~\mathrm{d}x - 6 \int x^{2} ~\mathrm{d}x + 2 \int x ~\mathrm{d}x + 5 \int 1 ~\mathrm{d}x.\end{align} \]
Apply the power rule for integrals\[ \begin{align}\int &\left( 4x^{3} - 6x^{2} + 2x + 5 \right) ~\mathrm{d}x \\&= \frac{4}{4}x^{4} - \frac{6}{3}x^{3} + \frac{2}{2}x^{2} + 5x \\&= x^{4} - 2x^{3} + x^{2} + 5x\end{align} \]
Add the integration constant.
\[ \begin{align}\int &\left( 4x^{3} - 6x^{2} + 2x + 5 \right) ~\mathrm{d}x \\&= x^{4} - 2x^{3} + x^{2} + 5x + C\end{align} \]
Verify your result by proving that \( F'(x) = f(x) \).\[ \begin{align}f(x) &= 4x^{3} - 6x^{2} + 2x + 5 \\F(x) &= x^{4} - 2x^{3} + x^{2} + 5x + C \\~\\F'(x) &= \left( x^{4} - 2x^{3} + x^{2} + 5x + C \right) \\&= 4x^{3} - 6x^{2} + 2x + 5 ~\checkmark\end{align} \]
This example requires you to simplify the integrand first.
Evaluate
\[ \int \frac{x^{2}+4\sqrt[3]{x}}{x} ~\mathrm{d}x \]
Solution:
Determine which properties and rules apply.
To better determine which rules to use, first split the fraction in the integrand:\[ \int \left( \frac{x^{2}}{x} + \frac{4\sqrt[3]{x}}{x} \right) ~\mathrm{d}x. \]
Now you can evaluate the integral term-by-term using the sum/difference rule and power rule.
If you need to use more than one property or rule, decide the order in which to use them.
Use the rules you decided on.
Apply the sum/difference rule.\[ \int \left( x + \frac{4}{x^{\frac{2}{3}}} \right) ~\mathrm{d}x = \int x ~\mathrm{d}x + 4 \int x^{-\frac{2}{3}} ~\mathrm{d}x \]
Apply the power rule.\[ \begin{align}\int \left( x + \frac{4}{x^{\frac{2}{3}}} \right) ~\mathrm{d}x &= \frac{1}{2}x^{2} + 4 \frac{1}{\left(\frac{-2}{3}\right)+1} x^{-\frac{2}{3}+1} \\&= \frac{1}{2}x^{2} + \frac{4}{\frac{1}{3}} x^{\frac{1}{3}} \\&= \frac{1}{2}x^{2} + 12x^{\frac{1}{3}}\end{align} \]
Add the integration constant.
\[ \int \left( x + \frac{4}{x^{\frac{2}{3}}} \right) ~\mathrm{d}x = \frac{1}{2}x^{2} + 12x^{\frac{1}{3}} + C \]
Verify your result by proving that \( F'(x) = f(x) \).\[ \begin{align}f(x) &= \frac{x^{2}+4\sqrt[3]{x}}{x} = \frac{x^{2}}{x} + \frac{4\sqrt[3]{x}}{x} = x + \frac{4}{x^{\frac{2}{3}}} \\F(x) &= \frac{1}{2}x^{2} + 12x^{\frac{1}{3}} + C \\~\\F'(x) &= \frac{2}{2}x + \frac{12}{3}x^{-\frac{2}{3}} \\&= x + 4x^{-\frac{2}{3}} \\&= x + \frac{4}{x^{\frac{2}{3}}} ~\checkmark\end{align} \]
This example calls on you to remember what the integrals of trigonometric functions look like.
Evaluate
\[ \int \frac{4}{1+x^{2}} ~\mathrm{d}x \]
Solution:
Determine which properties and rules apply.
If you need to use more than one property or rule, decide the order in which to use them.
Use the rules you decided on.
Apply the constant multiple rule.\[ \int \frac{4}{1+x^{2}} ~\mathrm{d}x = 4 \int \frac{1}{1+x^{2}} ~\mathrm{d}x \]
Apply the inverse tangent rule.\[ \int \frac{4}{1+x^{2}} ~\mathrm{d}x = 4 \tan^{-1}(x) \]
Add the integration constant.
\[ \int \frac{4}{1+x^{2}} ~\mathrm{d}x = 4 \tan^{-1}(x) + C \]
Verify your result by proving that \( F'(x) = f(x) \).\[ \begin{align}f(x) &= \frac{4}{1+x^{2}} \\F(x) &= 4 \tan^{-1}(x) + C \\~\\F'(x) &= 4 \cdot \frac{1}{1+x^{2}} \\&= \frac{4}{1+x^{2}} ~\checkmark\end{align} \]
This example shows that simplifying the trigonometric functions in the integrand can drastically simplify the problem.
Evaluate
\[ \int \tan(x) \cos(x) ~\mathrm{d}x \]
Solution:
Determine which properties and rules apply.
Rewrite the integrand as:\[ \begin{align}\int \tan(x) \cos(x) ~\mathrm{d}x &= \int \frac{\sin(x)}{\cancel{\cos(x)}} \cancel{\cos(x)} ~\mathrm{d}x \\&= \int \sin(x) ~\mathrm{d}x\end{align} \]
Now you know you just need to use the sine rule.
If you need to use more than one property or rule, decide the order in which to use them.
Use the rules you decided on.
Add the integration constant.
\[ \int \sin(x) ~\mathrm{d}x = -\cos(x) + C \]
Verify your result by proving that \( F'(x) = f(x) \).\[ \begin{align}f(x) &= \tan(x) \cos(x) = \frac{\sin(x)}{\cancel{\cos(x)}} \cancel{\cos(x)} = \sin(x) \\F(x) &= -\cos(x) + C \\~\\F'(x) &= -(-\sin(x)) \\&= \sin(x) ~\checkmark\end{align} \]
Indefinite Integral – Key takeaways
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