Moving backwards can be just as important as moving forwards, at least for math. Every operation or function in math has an opposite, usually called an inverse, used for “undoing” that operation or function. Adding has subtracting, squaring has square rooting, exponents have logarithms. Derivatives are no exception to this rule. If you can move forward to take a derivative, you can also move backwards to “undo” that derivative. This is called finding the antiderivative.
Antiderivative Meaning
For the most part, you need to know how to find antiderivatives for the process of integration. To explore integration further, see this article on Integrals.
The antiderivative of a function \(f\) is any function \(F\) such that \[F'(x)=f(x).\]
Note that Antiderivatives are usually notated using the capital letter version of the function name (that is, the antiderivative of \(f\) is \(F\) as shown in the definition).
Essentially, the antiderivative is a function that gives you your current function as a derivative.
In order to find an antiderivative, you need to know your differentiation rules very well. For some reminders about common differentiation rules, check out these articles on Differentiation Rules and Derivatives of Special Functions or see the table below under "Antiderivative Rules".
For example, if you have the function \(f(x)=2x\) and you need to find the antiderivative, you should ask yourself, "What function would give this result as a derivative?" You are probably familiar enough with finding derivatives at this point to know that \[\frac{d}{dx}(x^2)=2x.\] So, an antiderivative of \(f(x)=2x\) is \[F(x)=x^2.\]
You may also recognize the function \(F(x)=x^2\) is not the only function that will give you a derivative of \(f(x)=2x\). The function \(F(x)=x^2+5\), for example, would give you the same derivative and is also an antiderivative. Since the derivative of any constant is \(0\), there are infinitely many antiderivatives of \(f(x)=x^2\) of the form \[F(x)=x^2+C.\]
Antiderivative vs Integral
Antiderivatives and integrals are often conflated. And with good reason. Antiderivatives play an important role in integration. But there are some differences.
Integrals can be divided into two groups: indefinite integrals and definite integrals.
Definite integrals have bounds called bounds of integration. The purpose of a definite integral is to find the area under the curve for a specific domain. So, a definite integral will be equal to a single value. The general form for a definite integral will look something like, \[\int_a^b f(x)dx.\]
The variables \(a\) and \(b\) will be domain values, and you will be finding the area under the curve \(f(x)\) between those values.
The graph below shows an example of a definite integral. The function in consideration here is \(f(x)=x^2-2\), and the shaded region represents the definite integral \(\int_{-1}^{1} x^2-2 dx\).
Fig. 1. Example of the shaded region represented by a definite integral.
Indefinite integrals do not have bounds and are not limited to a particular interval of the graph. They also need to take into consideration the fact that any given function has infinitely many antiderivatives due to the possibility of a constant being added or subtracted. To show that there are many possibilities for an antiderivative, usually a constant variable \(C\) is added, like so,
\[\int f(x)dx=F(x)+C.\]
This allows you to denote the entire family of functions that could give you \(f(x)\) after differentiation and could therefore be antiderivatives.
For the example graph shown above of the function \(f(x)=x^2-2\), all the possible antiderivatives are \(F(x)=\frac{1}{3}x^3-2x+c\). The value \(C\) is called the constant of integration. Below shows a few different possible functions that \(F\) could be by changing the constant of integration.
Fig. 2. Graphs of some antiderivatives of \(f(x)=x^2-2.\)
If you need to take it a step further and solve for \(C\) in order to find a specific antiderivative function, see the article on Antiderivatives Initial Value Problems.
Antiderivative Formula
Considering again that the definition of an antiderivative is any function \(F\) that gives you your function \(f\) as a result of differentiation, you may realize that that means there will not be one formula for finding every antiderivative. At this point, you have learned many different rules for differentiating many different types of functions (power function, trig functions, exponential functions, logarithmic functions, etc.). Therefore, if you are finding the antiderivative of different types of functions, there will be a variety of rules. But the general idea for finding an antiderivative is to reverse the differentiation steps that you know. See the chart below in the next section, for specific antiderivative formulas for finding the antiderivative of common functions.
Properties of Antiderivatives
There are some properties that may make it easier to find antiderivatives for some functions. The Sum Rule and The Difference Rule (explained in the article on Differentiation Rules) both apply to antiderivatives as they do to derivatives.
Recall that differentiation is linear, which means that the derivative of a sum of terms is equal to the sum of the derivatives of the individual terms, and the derivative of a difference of terms is equal to the difference of the derivatives of the individual terms.
Integration is also linear. The antiderivative of the sum of multiple terms is equal to the sum of the antiderivatives of the individual terms, the same applies for \[\int f(x) \pm g(x) dx=\int f(x)dx\pm\int g(x)dx=F(x)\pm G(x)+C.\]
The Constant Multiple Rule also applies to antiderivatives. The antiderivative of a function that is multiplied by a constant \(k\) is equal to the constant \(k\) multiplied by the antiderivative of the function. You can essentially "factor out" a constant from the integral before finding the antiderivative, \[\int k\cdot f(x)dx=k\int f(x)dx=kF(x)+C.\]
Mistakes to Avoid
As is the case with most things in math, the rules that apply to addition and subtraction do not apply in the same measure to multiplication and division. So, there is no property saying that the antiderivative of the product or quotient of two functions would be the same as the product or quotient of the antiderivatives of the functions, \[\int f(x)\cdot g(x)dx \neq \int f(x)dx \cdot \int g(x)dx.\]
Finding antiderivatives for these kinds of functions will be much more involved. Recall that the Product Rule for differentiation is, \[\frac{d}{dx}(f(x)\cdot g(x))=f(x)\frac{dg}{dx}+g(x)\frac{df}{dx}.\]
So finding antiderivatives of functions with products in them means that either a chain rule was applied during differentiation or the product rule was used. To tackle antiderivatives like these, you can check out the articles on Integration by Substitution and Integration by Parts.
Antiderivative Rules
The rules for finding antiderivatives are generally the reverse of the rules for finding derivatives. Below is a chart showing common antiderivative rules.
| Differentiation Rule | Associated Antiderivative Rule |
| The Constant Rule. \(\dfrac{d}{dx}(C)=0.\) | \(\int 0dx=C.\) |
| The Power Rule. \(\dfrac{d}{dx}(x^n)=nx^{n-1}.\) | \(\int x^ndx=\dfrac{x^{n+1}}{n+1}+C, n \neq -1.\) |
| The Exponential Rule (with \(e\)). \(\dfrac{d}{dx}(e^x)=e^x.\) | \(\int e^xdx=e^x+C.\) |
| The Exponential Rule (with any base \(a\)). \(\dfrac{d}{dx}(a^x)=a^x \cdot \ln a.\) | \(\int a^xdx=\dfrac{a^x}{\ln a}+C, a \neq 1.\) |
| The Natural Log Rule. \(\dfrac{d}{dx}(\ln x)=\dfrac{1}{x}.\) | \(\int \dfrac{1}{x}dx=\ln |x|+C.\) |
| The Sine Rule. \(\dfrac{d}{dx}(\sin x)=\cos x.\) | \(\int \cos xdx=\sin x + C.\) |
| The Cosine Rule. \(\dfrac{d}{dx}(\cos x)=-\sin x.\) | \(\int \sin xdx=-\cos x +C.\) |
| The Tangent Rule. \(\dfrac{d}{dx}(\tan x)= \sec^2 x.\) | \(\int \sec^2 xdx=\tan x + C.\) |
| The Cotangent Rule. \(\dfrac{d}{dx}(\cot x)=-\csc^2 x.\) | \(\int \csc^2 xdx=-\cot x + C.\) |
| The Secant Rule. \(\dfrac{d}{dx}(\sec x)=\sec x \tan x.\) | \(\int \sec x \tan xdx=\sec x + C.\) |
| The Cosecant Rule. \(\dfrac{d}{dx}(\csc x)=-\csc x \cot x.\) | \(\int \csc x \cot x dx =-\csc x + C.\) |
Table 1. Differentiation rules and their antiderivatives.
Antiderivative Examples
Let's look at a few examples that use the rules outlined above.
Let's say that you are given a function that describes a particle's velocity, \(f(x)=x^3-10x+8\) where \(x\) is the time in seconds of the particle's movement. Find all possible position functions for the particle.
Solution:
First, recall that velocity is the derivative of position. So in order to find the position function \(F\), you need to find the antiderivatives of the velocity function \(f\) you are given, \[\int 3x^2-10x+8dx=F(x).\]
For this antiderivative, you can start by using both the sum rule and the constant multiple rule to individualize the terms. Then you can use the Power Rule on each term to find the antiderivative of each individual term,
\[\begin{align} \int 3x^2-10x+8dx&=3\int x^2dx-10\int xdx+\int 8dx+C.\\&=3\left(\frac{x^3}{3}\right)-10\left(\frac{x^2}{2}\right)+8x+C.\\\int 3x^2-10x+8dx&=x^3-5x^2+8x+C.\\\end{align}\]
Thus, all possible position functions for \(f\) are \[F(x)=x^3-5x^2+8x+C.\]
Your next steps from here would depend on the type of problem you are being asked to solve. You could be asked to find a specific position function by doing an initial value problem. Or you may be asked to how far the particle traveled over a specific interval of time by solving a definite integral problem.
Now let's look at an example that demonstrates how important it is to recognize your derivative rules.
Find all possible antiderivatives \(F\) for the function \(f(x)=\dfrac{5}{4x}\).
Solution:
First, you will use the constant multiple rule to factor out the coefficients in both the numerator and the denominator. This really cleans up the problem so that it will be easier to recognize which derivative rule you are looking for, \[F(x)=\int \frac{5}{4x}dx=\frac{5}{4} \int \frac{1}{x}dx.\]
If you do not immediately recognize which antidifferentiation rule to apply here, you may try to reverse the Power Rule since it often works when the variable has negative and/or fractional exponents. But you will quickly run into the problem of getting \(x^0\) after adding 1 to the power. This is of course a problem since \(x^0=1\) and then \(x\) would disappear! So think back to your differentiation rules to remember when you got a derivative of \(\frac{1}{x}\) as a result. This is the derivative for \(\ln x\). So you can now use that to find the antiderivatives,
\[\begin{align} F(x)&=\frac{5}{4} \int \frac{1}{x}dx.\\&=\frac{5}{4} (\ln |x|)+C.\\F(x)&=\frac{5}{4} \ln |x| + C \,\,\,\text{ or }\,\,\, F(x)=\ln |x|^{\frac{5}{4}}+C.\\ \end{align}\]
The final example can be a tricky one. Notice that the antiderivative table above does not have the antiderivative of \(\tan x\). Seems like it should be a pretty simple antiderivative to find, doesn't it? Well, it is not quite as straightforward as its sine and cosine counterparts. It requires knowing your trigonometric properties and integration by substitution.
Find the general antiderivative of \(f(x)=\tan x\).
Solution:
Because tangent is not the direct result of any of the differentiation rules, you will need to try something different for it. Start by rewriting tangent using the trig properties you know,
\[\int \tan xdx=\int \frac{\sin x}{\cos x} dx.\]
This ends up being pretty helpful because the derivative of sine is cosine and the derivative of cosine is negative sine. You will use this fact to make a \(u\)-substitution. Here we will choose cosine for \(u\),
\[\begin{align} u&=\cos x.\\ du&=-\sin xdx.\\ -du&=\sin xdx.\\ \end{align}\]
Now make your substitution, \[\int \tan xdx=-\int \frac{1}{u}du.\]
You can see here that this looks like the derivative rule for natural log:
\[\begin{align} \int \tan xdx&=-\int \frac{1}{u}du.\\ \int \tan xdx&=-\ln |u| + C.\\ \end{align}\]
Now you can substitute back in for u,
\[\int \tan xdx=-\ln |cos x| +C.\]
As it turns out, tangent is a simple function with a not so simple antiderivative.
Antiderivative of Inverse Trig Functions
Inverse trig functions are kind of a strange case when it comes to both differentiation and integration. The derivatives of inverse trig functions don't really look like they would be related to the inverse trig functions themselves. You should be on the lookout for Integrals Resulting in Inverse Trigonometric Functions (explored here in more depth). For a reminder, below is a table showing the differentiation rules for the inverse trig functions and the associated antiderivatives:
| Differentiation Rule | Associated Antiderivative |
| The Arcsine Rule. \(\dfrac{d}{dx}(\sin ^{-1}x)=\dfrac{1}{\sqrt{1-x^2}}.\) | \(\int \dfrac{1}{\sqrt{1-x^2}}dx=\sin^{-1}x+C.\) |
| The Arccosine Rule. \(\dfrac{d}{dx}(\cos^{-1}x)=\dfrac{-1}{\sqrt{1-x^2}}.\) | \(\int \dfrac{-1}{\sqrt{1-x^2}}dx=\cos^{-1}x+C.\) |
| The Arctangent Rule. \(\dfrac{d}{dx}(\tan^{-1}x)=\dfrac{1}{1+x^2}.\) | \(\int \dfrac{1}{1+x^2}dx=\tan^{-1}x+C.\) |
| The Arcsecant Rule. \(\dfrac{d}{dx}(\sec^{-1}x)=\dfrac{1}{|x|\sqrt{x^2-1}}.\) | \(\int \dfrac{1}{|x|\sqrt{x^2-1}}dx=\sec^{-1}x+C.\) |
| The Arccosecant Rule. \(\dfrac{d}{dx}(\csc^{-1}x)=\dfrac{-1}{|x|\sqrt{x^2-1}}.\) | \(\int \dfrac{-1}{|x|\sqrt{x^2-1}}dx=\csc^{-1}x+C.\) |
| The Arccotangent Rule. \(\dfrac{d}{dx}(\cot^{-1}x)=\dfrac{-1}{1+x^2}.\) | \(\int \dfrac{-1}{1+x^2}dx=\cot^{-1}x+C.\) |
Table 2. Differentiation rules for inverse trigonometric functions and their antiderivatives.
The antiderivatives of inverse trig functions have a lot going on (but at least look a little bit more related). Below is a chart of the antiderivatives of inverse trig functions. They are achieved by using the methods Integration by Parts and Integration by Substitution:
Table 3. Differentiation rules for inverse trigonometric functions and their antiderivatives.
| Inverse Trig Function | Antiderivatives of Inverse Trig Functions |
| Arcsine Antiderivative. | \(\int \sin^{-1}xdx=x\sin^{-1} x + \sqrt{1-x^2}+C.\) |
| Arccosine Antiderivative. | \(\int \cos^{-1} xdx=x\cos^{-1} x - \sqrt{1-x^2}+C.\) |
| Arctangent Antiderivative. | \(\int \tan^{-1} xdx=x\tan^{-1} x - \frac{1}{2} \ln |1+x^2|+C.\) |
| Arcsecant Antiderivative. | \(\int \sec^{-1} xdx=x\sec^{-1} x - \ln |x+\sqrt{x^2-1}|+C.\) |
| Arccosecent Antiderivative. | \(\int \csc^{-1} xdx=x\csc^{-1} x + \ln |x+\sqrt{x^2-1}|+C.\) |
| Arccotangent Antiderivative. | \(\int \cot^{-1} xdx=x\cot^{-1}x + \frac{1}{2} \ln |1+x^2|+C.\) |
You may be wondering where in the world those antiderivatives for the inverse trig functions come from. Below, we will walk through the process of finding the antiderivative of the arcsine function. The process uses both Integration by Parts and Integration by Substitution, so be sure you are familiar with those first.
We will begin with Integration by parts, which means that our function will need to be split into two parts, \[\int \sin^{-1} xdx=\int \sin^{-1} x \cdot 1dx.\]
Now recall that integration by parts \[\int udv=uv-\int vdu\] so we now need to choose our parts. One part will be assigned as \(u\) and the other part assigned as \(dv\). Using the LIATE rule of thumb (outlined in the integration by parts article), we will choose \(u\) to be the inverse trig function. Once \(u\) and \(dv\) are assigned, we also need to find \(du\) and \(v\), like so:
| \(u=sin^{-1}x.\) | \(v=x.\) |
| \(du=\frac{1}{\sqrt{1-x^2}}dx.\) | \(dv=1dx.\) |
Now we can substitute in each part:
\[\begin{align} \int udv&=uv-\int vdu.\\ \int \sin^{-1}x \cdot 1dx&=x\sin^{-1} x - \int \frac{x}{\sqrt{1-x^2}}dx.\\ \end{align}\]
Now we need to focus on the last term, which is a new integral. To find the antiderivative of the second integral, we will have to use integration by substitution, also known as \(u\)-substitution. For this, we will choose that,
\[\begin{align} u&=1-x^2.\\ du&=-2xdx.\\ -\frac{1}{2}du&=xdx.\\ \end{align}\]
Next, we will pick up where we left off, but focusing on integrating the last term using the \(u\)-substitution chosen above,
\[\begin{align} \int \sin^{-1}xdx&=x\sin^{-1}x-\int \frac{x}{\sqrt{1-x^2}}dx.\\&=x\sin^{-1}x-\int -\frac{1}{2} \cdot \frac{1}{\sqrt{u}}du.\\&=x\sin^{-1}x+ \frac{1}{2}\int \frac{1}{\sqrt{u}}du.\\&=x\sin^{-1}x+\frac{1}{2}\int u^{-\frac{1}{2}}du.\\\end{align}\]
At this point, in order to integrate, we need to use the power rule,
\[\begin{align} \int \sin^{-1}xdx&=x\sin^{-1}x+\frac{1}{2} \left(\frac{u^{\frac{1}{2}}}{\frac{1}{2}}\right)+C.\\&=x\sin^{-1}x+u^{\frac{1}{2}}+C.\\&=x\sin^{-1}x+\sqrt{u}+C.\\\end{align}\]
And finally, substitute back in for \(u\) to get your final antiderivative, \[\int \sin^{-1}xdx=x\sin^{-1}x+\sqrt{1-x^2}+C.\]
The steps for finding the other inverse trig functions' antiderivatives will be similar, and you will need to employ similar strategies.
Antiderivatives - Key takeaways
- An antiderivative of \(f\) is a function \(F\) such that \(F’(x)=f(x).\) It is a way to “undo” differentiation.
- There are infinitely many antiderivatives for any given function, so the antiderivative family of functions will often be written as an indefinite integral defined as \(\int f(x)=F(x)+C\).
- There is no one formula for finding the antiderivative. There are many basic formulas for finding antiderivatives of common functions based on common differentiation rules.
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