# Indefinite Integrals

Now that you’ve got an understanding of what derivatives are and how to find them, we are going to move on to one of the last major topics of Calculus I: antiderivatives.

As the name suggests, an antiderivative is essentially the undoing of a derivative. For example, given some function $$f(x)$$, its antiderivative—which is usually called $$F(x)$$—is the function that, when you take its derivative, will give you $$f(x)$$.

$$F(x)$$ is the antiderivative of $$f(x)$$ if $$F'(x)=f(x)$$

This may sound a little funky, but let’s try it out with an example. Find the antiderivative of $$f(x)=2x$$.

In order to find the antiderivative, we need to think of $$2x$$ as being a derivative already. What function would give us $$2x$$ as its derivative? We know that from experience that $$x^{2}$$ would! Because of the Power Rule, the derivative of $$x^{2}$$ is $$2x$$. Let’s go ahead and write that $$F(x)$$ is $$x^{2}$$.

$$F(x)=x^{2}$$ …

Antiderivatives are often as simple as doing a reversal of the Power Rule—but with one catch. We said that the derivative of $$x^{2}$$ is $$2x$$. But what about the derivative of $$x^{2}+1$$? That would also make $$2x$$. The same goes for $$x^{2}+2$$, or $$x^{2}+3$$, or $$x^{2}$$ plus or minus any constant. Because of the Constant Rule, the derivative of these second terms will always be zero, and the resulting derivative comes out to $$2x$$.

$$\frac{d}{dx}(x^{2}+1)=2x$$

$$\frac{d}{dx}(x^{2}+2)=2x$$

$$\frac{d}{dx}(x^{2}+3)=2x$$

$$\frac{d}{dx}(x^{2}+c)=2x$$

Because of this, when we take an antiderivative, we must add an arbitrary constant to the answer. That way we are covered for every possible one of these solutions. In our example, we would write that $$F(x)=x(2)+C$$. That $$C$$ is the arbitrary constant we need to protect our answer from missing solutions.

Let’s try another example. Find the antiderivative of $$g(x)=5x^{4}+7x$$.

This problem, like the first, requires a reversal of the Power Rule. Since the Power Rule tells us that we can find a derivative by multiplying by the power of $$x$$ and then reducing the exponent by 1, we are going to do the opposite of these things, and in the opposite order. We need to add 1 to the exponent, and then divide by the new power.

For the first term, $$5x^{4}$$, we add 1 to the exponent, making $$5x^{5}$$, then divide by the new power. Dividing by 5 gives us $$x^{5}$$.

For the second term, we increase the power by 1 to get $$7x^{2}$$, then divide by the value of the new power to get $$\frac{7x^{2}}{2}$$.

The antiderivative for this problem is then $$G(x)=x^{5}+\frac{7x^{2}}{2}$$… plus $$C$$. Don’t forget to add the constant! We can check this answer by taking the derivative . Using the Power Rule, the first term would go back to $$5x^{4}$$, the second term would go back to $$7x$$, and the $$C$$ would go to zero. This is exactly what we started the problem with!

$$G(x)=x^{5}+\frac{7x^{2}}{2}+C$$

$$G'(x)=5x^{4}+7x=g(x)$$

Antiderivatives are more formally referred to as “indefinite integrals.” As we have talked about already, antiderivatives can be identified by their capital letter notation. However, moving forward, you’ll probably come across integral notation more prominently. This means that instead of using the words “find the antiderivative of this function,” these indefinite integral problems are shortened by instead asking for the integral

$$\int$$

…of that function…

$$\int f(x)$$

…with respect to $$x$$.

$$\int f(x) dx$$

The long “S” shape is what we call the integral symbol, while the letters $$dx$$ specify that we are taking the integral with respect to the variable $$x$$. Every once in a while you may have a problem in terms of some other variable, but they can be solved in the same way. As we have discussed, the antiderivative, or indefinite integral, of some function $$f(x)$$ with respect to $$x$$, is $$F(x)$$ plus a constant $$C$$.

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

First, the integral of $$sin(x)$$ is $$-cos(x)+C$$.

$$\int sin(x) dx=-cos(x)+C$$

This should feel familiar because we know that the derivative of $$-cos(x)$$ is $$sin(x)$$. Likewise, the integral of $$cos(x)$$ is familiar; we know it must be $$sin(x)+C$$.

$$\int cos(x) dx=sin(x)+C$$

The next integral I want you to write down is the integral of $$sec^{2}(x)$$, which is $$tan(x)+C$$.

$$\int sec^{2}(x) dx=tan(x)+C$$

Next, the integral of $$csc^{2}(x)$$ is $$-cot(x)+C$$.

$$\int csc^{2}(x) dx=-cot(x)+C$$

The integral of $$sec(x)tan(x)$$ is $$sec(x)$$ plus a constant $$C$$.

$$\int sec(x)tan(x) dx=sec(x)+C$$

The last trig integral I have for you is the integral of $$csc(x)cot(x)$$, which is $$-csc(x)+C$$.

$$\int csc(x)cot(x) dx=-csc(x)+C$$

Since the exponential function $$e^{x}$$ is its own derivative, it should come as no surprise that it is also its own integral!

$$\int e^{x} dx=e^{x}+C$$

And finally, the integral of $$\frac{1}{x}$$ is  $$ln(x)+C$$.

$$\int \frac{1}{x} dx= ln(x)+C$$

I know that this seems like a lot of information, but if you spend a few minutes writing each of these out a few times, you’ll start to become familiar with them and memorization won’t be far from reach. You can do this after you finish the video, or you can pause the video and do it now.

Let’s close by working through one more example. Determine the indefinite integral of $$h(x)=8x^{3}+sec(x)tan(x)+e^{x}$$.

$$\int (8x^{3}+sec(x)tan(x)+e^{x}) dx$$

The Sum and Difference Rule told us that we can take the derivative of a function one term at a time. The same rule applies to integrals! We can simplify this expression by splitting up the terms into separate integrals.

$$\int 8x^{3} dx+sec(x)tan(x) dx+e^{x} dx$$

Even though this line is a little longer than the first, we can more easily see how to handle each part now. First, $$8x^{3}$$ can be integrated by reversing the Power Rule. Let’s add 1 to the exponent to get $$8x^{4}$$, then divide by the value of this new power to get $$\frac{8x^{4}}{4}=2x^{4}$$. We can always double-check this by taking the derivative again. The derivative of $$2x^{4}$$ is indeed $$8x^{3}$$. Don’t forget to write the “plus $$C$$.” Since we have three integrals in this problem, I’m going to write a little “1” to indicate that this constant came from the first part.

$$2x^{4}+C_{1} +$$ …

The next term, $$sec(x)tan(x)$$, is one that we discussed moments ago. Did you write it down? Its integral is $$sec(x)$$ plus a constant. I’ll write a little “2” with this $$C$$ to show that it came from the second integral.

$$2x^{4}+C_{1}+sec(x)+C_{2}+$$…

The third term should integrate quickly. As we said, the integral of $$e^{x}$$ is $$e^{x}$$, so let’s write that plus a third constant.

$$2x^{4}+C_{1}+sec(x)+C_{2}+e^{x}+C_{3}$$

Now, let’s clean this solution up just a little bit. Because $$C_{1}$$, $$C_{2}$$, and $$C_{3}$$ are all constants, they’re all just numbers. And their sum is just going to be another number. For that reason, we can combine the $$C$$s into one constant together. So the final answer for this problem is:

$$2x^{4}+sec(x)+e^{x}+C$$

Remember, when integrating $$x$$ to some power, you can reverse the Power Rule by adding 1 to the exponent and then dividing by the new power. Integrals of trig functions will require a little memorization, but you can get a good grip on them within just a few minutes of writing. The exponential function $$e^x$$ is unchanged by integration, and the integral of $$\frac{1}{x}$$ is $$ln(x)$$. And in all indefinite integral problems, it is critical to remember your “$$+C$$”!

I hope this video was helpful. Thanks for watching, and happy studying!

## Indefinite Integral Practice Questions

Question #1:

Evaluate the indefinite integral $$\int365 dx$$.

$$365x+c$$
$$365+c$$
$$3x+6x+5x+c$$
$$365x^2+c$$

Remember, one property of integrals is that any integral of the form $$\int k\text{ }dx=kx+c$$, where $$k$$ and $$c$$ are both constants. In this problem, $$365$$ is in the place of the constant $$k$$, so to get the solution, simply multiply it by $$x$$ and add the integration constant $$c$$. The result is $$365x+c$$, which can be checked by taking its derivative: $$\frac{d}{dx}(365x+c)=365$$.

Question #2:

Evaluate the indefinite integral $$\int100x^{99} dx$$.

$$x^{100}$$
$$x^{100}+c$$
$$x^{99}+100$$
$$x^{101}+c$$

For this problem, the following formula is especially helpful: $$\int x^n dx=\frac{x^{n+1}}{n+1}$$. In this problem, $$n=99$$, so the solution can be found by increasing that exponent by $$1$$, dividing by $$n+1=100$$, and finally adding the integration constant $$c$$.
$$\int100x^{99} dx=\frac{100}{99+1}x^{99+1}+c=x^{100}+c$$

This solution can be checked by taking its derivative:
$$\frac{d}{dx}(x^{100}+c)=100x^{99}$$

Question #3:

Evaluate the indefinite integral $$\int(x^4+x^3+x^2+x+1) dx$$.

$$x^5+x^4+x^3+x^2+x+c$$
$$\frac{1}{4}x^5+\frac{1}{3}x^4+\frac{1}{2}x^3+x^2+\frac{1}{x}+c$$
$$4x^3+3x^2+2x+1+c$$
$$\frac{1}{5}x^5+\frac{1}{4}x^4+\frac{1}{3}x^3+\frac{1}{2}x^2+x+c$$

To get this solution, work with each term individually. With each term, increase the exponent by $$1$$ and divide by the new exponent. It is important to remember the integration constant $$c$$, but this can be added at the very end.
$$\int x^4 dx=\frac{1}{5}x^5$$
$$\int x^3 dx=\frac{1}{4}x^4$$
$$\int x^2 dx=\frac{1}{3}x^3$$
$$\int x dx=\frac{1}{2}x^2$$
$$\int 1 dx=x$$

$$\int (x^4+x^3+x^2+x+1) dx=\frac{1}{5}x^5+\frac{1}{4}x^4+\frac{1}{3}x^3+\frac{1}{2}x^2+x+c$$

This solution can be checked by taking its derivative:
$$\frac{d}{dx}(\frac{1}{5}x^5+\frac{1}{4}x^4+\frac{1}{3}x^3+\frac{1}{2}x^2+x+c)=x^4+x^3+x^2+x+1$$

Question #4:

Evaluate the indefinite integral $$\int(-\frac{3}{2}x^4+3x^2) dx$$.

$$-\frac{1}{5}x^5+3x^3+c$$
$$-3x^5+x^3+c$$
$$-\frac{3}{10}x^5+x^3+c$$
$$-6x^3+6x+c$$

First, notice that the integrand in this problem has multiple terms. Each term will be evaluated separately, beginning with the first one: $$\int -\frac{3}{2}x^4 dx$$.

The fraction here can get in the way, so apply the property that constants can be factored out to move $$-\frac{3}{2}$$ outside the integral. The term is then $$-\frac{3}{2}\int x^4 dx$$, which is easier to solve. Increase the exponent by $$1$$, then divide by that new exponent.
$$-\frac{3}{2}\int x^4 dx=-\frac{3}{2}\times\frac{1}{5}x^5=-\frac{3}{10}x^5$$

The second term of the integrand can be solved by again using the rule of increasing the exponent and dividing.
$$\int 3x^2 dx=\frac{3}{2+1}x^{2+1}=x^3$$

The solution of the entire integral can be found by putting these two pieces together and adding the integration constant $$c$$:
$$\int (-\frac{3}{2}x^4+3x^2 ) dx=-\frac{3}{10}x^5+x^3+c$$

Question #5:

Caleb went to a drag race recently to study the speed and displacement of the cars. He determined that the speed of one car during the race could be expressed as the function $$f(x)=100x$$, where $$x$$ represents time in seconds. Determine $$F(x)$$, the function of the car’s displacement over time $$x$$, by taking the integral of $$f(x)$$.

$$F(x)=50x^2+c$$
$$F(x)=50x^2$$
$$F(x)=100+c$$
$$F(x)=50x^2+100x+c$$

Velocity and displacement have a fascinating relationship because displacement can be found by taking the integral of velocity. Conversely, velocity can be found by taking the derivative of displacement. In this problem, we find the function of displacement, $$F(x)$$, by taking the integral of the function of velocity, $$f(x)$$. In other words, simply evaluate the integral $$\int f(x)dx=\int 100x dx$$. First, raise the power of $$x$$ by $$1$$, then divide by the new exponent. Finally, add the integration constant $$c$$.
$$\int 100x dx=\frac{100}{1+1}x^{1+1}+c=50x^2+c$$

So $$F(x)=50x^2+c$$ represents the position, or displacement, of the car during the race. This solution can be checked by taking its derivative:
$$\frac{d}{dx}(50x^2+c)=100x$$