An antiderivative, as the name suggests, is just going backwards from taking a derivative. That is, instead of finding the derivative function, we find a function whose derivative is our starting function. In order to be able to find these antiderivatives, we will begin by looking at derivatives and derivative rules we know, and seeing what happens if we go backwards.

**All antiderivatives**

As we begin to look at antiderivatives, we often want to find all functions that when you take the derivative, you get the original function. When we do this, we say that we are finding the indefinite integral of \(f(x)\) with respect to \(x\), and we denote this as

\begin{align*}

\int f(x) dx.

\end{align*}

**Example**

The first thing we need to note is that since we want to find all antiderivatives, this can be a large task. In fact there can be infinitely many such functions. For example, let’s look at

\begin{align*}

\int x^{3}dx.

\end{align*}

To begin with, we note that \(\frac{d}{dx}\frac{1}{4}x^{4}=x^{3}\) so \(\frac{1}{4}x^{4}\) is one antiderivative of \(x^{3}\). However, \(\frac{d}{dx}\frac{1}{4}x^{4}+5=x^{3}\) as well, so this is also an antiderivative. If we generalize this, we see that \(\frac{d}{dx}\frac{1}{4}x^{4}+c=x^{3}\) for any constant \(c\).

**Are there more?**

Now, the next thing we would like to know is if there are any other functions that are antiderivatives. In order to see if this is possible, let’s assume that \(G(x)\) is a function that satisfies \(G'(x)=x^{3}\). If this is the case, then we see that

\begin{align*}

&\frac{d}{dx}(G(x)-\frac{1}{4}x^{4}) \\

&=\frac{d}{dx}G(x)-\frac{d}{dx}\frac{1}{4}x^{4} \\

&=x^{3}-x^{3} \\

&=0.

\end{align*}

That is, the the derivative of the function \(G(x)-\frac{1}{4}x^{4}\) is \(0\).

Now recall that the mean value theorem tells us that for any function, \(f(x)\) that is continuous on \([a,b]\) and differentiable on \(a,b)\), we have that

\begin{align*}

\frac{f(b)-f(a)}{b-a}=f'(c)

\end{align*}

for some \(c\) in \((a,b)\). Since the derivative of \(G(x)-\frac{1}{4}x^{4}=0\), it is continuous and differentiable on the interval. If we let \(f(x)=G(x)=\frac{1}{4}x^{4}\), we get that, for any \(a,b\),

\begin{align*}

\frac{f(b)-f(a)}{b-a}&=0 \\

f(b)-f(a)&=0 \\

f(b)&=f(a).

\end{align*}

That is, \(f(x)\) is a continuous function, so \(f(x)=c\) for some constant \(c\). Hence

\begin{align*}

G(x)-\frac{1}{4}x^{4}&=c \\

G(x)&=\frac{1}{4}x^{4}+c.

\end{align*}

What we have found here is that if we can find one antiderivative of \(x^{3}\), we can find infinitely many by adding an arbitrary constant to the function. Furthermore, every antiderivative must differ from the one we found by only a constant, hence all antiderivative will be of this form. While we haven’t shown this in general, it will follow that this is true for all functions, not just \(x^{3}\). In this case, we find

\begin{align*}

\int x^{3}dx=\frac{1}{4}x^{4}+c.

\end{align*}

**More Indefinite Integrals**

Now that we’ve seen that if we want to find all antiderivatives, we really just need to find one and add a constant of integration, we are ready to find some indefinite integrals. Instead of going through the entire list one-by-one here, we will point out that we can use what we know about derivatives to show that each of the following integral rules is correct.

- \(\int f(x) \pm g(x) dx =\int f(x)dx \pm \int g(x)dx\),
- \(\int kf(x)dx=k\int f(x)dx\),
- \(\int x^{n}dx=\frac{1}{n+1}x^{n+1}+c \), if \(n \neq 0\)
- \(\int x^{-1}dx=\int \frac{1}{x}dx=\ln|x|+c\),
- \(\int e^{x}dx=e^{x}+c\),
- \(\int \sin(x)dx=-\cos(x)+c \),
- \(\int \cos(x)dx=\sin(x)+c\)

We can and will find more indefinite integrals in the future, but this is a good list to start with.

**Example**

Using the above rules and given indefinite integrals find

\begin{align*}

\int 2x^{3}-\frac{3}{x^{2}}-\frac{1}{x}+\sin(x)dx.

\end{align*}

As we begin this problem, we note that we don’t want to do everything at once. Instead, we will break this into pieces using the first two indefinite integrals rules we found above. That is, we can distribute indefinite integrals across sums, differences and factor out constants. We can now see that

\begin{align*}

&\int 2x^{3}-\frac{3}{x^{2}}-\frac{1}{x}+\sin(x)dx \\

&=2\int x^{3}dx-3\int\frac{1}{x^{2}}dx-\int\frac{1}{x}dx+\int\sin(x)dx.

\end{align*}

The first integral we can find using the third rule above. The second one; however, doesn’t seem to follow from the above rules. We do note that we can rewrite this so that instead of being written as a fraction, it can be written as a negative power function. This will again allow us to use the third rule. While it may look like we should do the same thing for the third integral, we instead have to use our fourth rule because \(x^{-1}\) is not included in the third rule. The last integral follows form our sixth rule. We now find,

\begin{align*}

&\int 2x^{3}-\frac{3}{x^{2}}-\frac{1}{x}+\sin(x)dx \\

&=2\int x^{3}dx-3\int\frac{1}{x^{2}}dx-\int\frac{1}{x}dx+\int\sin(x)dx\\

&=2\int x^{3}dx-3\int x^{-2}dx-\int\frac{1}{x}dx+\int\sin(x)dx\\

&=2\frac{1}{4}x^{4}-3\frac{1}{-1}x^{-1}-\ln|x|-\cos(x)+c.

\end{align*}

Remember that we need to the constant at the end so that we have all antiderivatives. Furthermore, you are welcome to simplify the solution if you would like. However, since we have a correct answer, I will leave the solution as is.

**Conclusion**

By noting that we can find all antiderivatives by finding one and adding a constant of integration, we have reduced finding integrals to a guess and check process. That is, we can make educated guesses using what we know about derivatives. If we check and see that our solutions is correct, we then need only add a constant of integration in order to find all antiderivatives. In this way, we can indeed find many indefinite integrals.

I hope you found this useful as you work on finding indefinite integrals. If you did, make sure to like and share the post. Also be sure to subscribe to our YouTube channel to get notified about the helpful videos we post.