Indefinite Integrals – Definitely Easier Now!

Hey guys! Welcome to this video on indefinite integrals.

At this point, I’m assuming you are familiar with derivatives or differentiation. Well, when working with integrals you keep the same principle of working with derivatives in mind, except you kind of work backwards.

So, let’s look at a function:

        f(x)= x^7+4x +11

Which function did we differentiate to get this function?

This question may seem harder than it is if you’ve never been asked. So, let’s take a logical approach.

First, let’s take the derivative of this function.

Recall, that when differentiating powers of x, we bring our original exponent down to multiply times the base, and subtract one from the exponent. When we do that with this function, we get:

        f’(x)= 7x^6+4+C

Alright, now the reason we just looked at how to take derivatives was 1) to refresh your memory, and 2) to make you think about what exactly is happening when we are taking the derivative.

So, now, let’s look back at our original question: Which function did we differentiate to get this function ( f(x)= x^7+4x +11 )?

Let’s work with what we know. We know that when this function was differentiated that 1 was subtracted from the exponent. So, math and logic tell us that if we add one to this exponent then we should get the original exponent we started with. What else do we know? Well, we know that whatever that original exponent is, which we know is 8, was brought down and multiplied by our base; but when we look at our function here we don’t have an 8 being multiplied, which means that whatever 8 was multiplied by caused it to cancel out. Well again, just use a little math and logic and we know that 8 times ⅛ cancel out.

So, where does that put us. We know that our original exponent was 8, so that would give us x^8, but we also know that when we took the derivative and brought 8 down to be multiplied it canceled out. So,we know that there is a ⅛ in the coefficient. So, our first term here is ⅛ x^8. Well, we do this same thing to each of our other two terms. Let’s look at our 4x here. Well, again, one was taken away from the exponent, so we need to add a 1. Which, gives us x^2. Now, the two in our exponent was brought down, and multiplied by something that gave us 4. Well, 2 times what is 4? 2! Which makes our original term, before the derivative was taken, 2x^2. We are almost done with this one. It looks like our last term here dropped it’s variable during the derivative process, and we know there had to have been a variable with 11 or else it was just a constant… and the derivative of a constant is zero. So, one was subtracted, which obviously left us with a zero in the exponent, because anything to the power of zero is 1. So, let’s add that one back. Which means we can just bring back the x. X has an exponent of one, and anything times one is itself. So, our 11 stays and we have 11x.

This process of finding the original function of a function that has been differentiated is what we call integration. Another way you could say this is finding the antiderivative or indefinite integral.

But here is a more formal definition:

The antiderivative of a derivative is the original function plus a constant.
So, if (d/dx)f(x)=g(x), then that means that g(x) is the derivative of f(x)
and that f(x) is the antiderivative of g(x)
∫g(x)dx= f(x) + C

Now, a couple things before we jump in to an example. First, when you have your integral sign here, you must always place your dx at the end. The dx let’s you know where the integrand ends.

Secondly, there are a few properties that you need to be aware of.

The first is:

      ∫kf(x)dx = kഽf(x)dx          where k is any constant.

What this properties tells us is that we can factor out any multiplicative constant.

The second property is:

     ഽ-f(x)dx=-ഽf(x)dx         where k is any negative constant.

This is really the first property, except it’s letting you know that you can factor out the negative sign as well.

The third and last property that we will talk about in this video is:

          ഽf(x) + g(x)dx = ഽf(x)dx + ഽg(x)dx

What this property tells us is that when terms are being added or subtracted we can take their individual integrals.

Now, let’s workout a few examples.

So, in the first example I want to start with the original function, take the derivative, then integrate. I won’t do this with every example, but I want you guys to see what happens with a constant, and to show you guys that we will actually get back our original function when we integrate.

       f(x)= 6x^3 + 8
ഽ18x^2dx =

Now, you already know the answer. When we integrate this function, we should get back to our original f(x) function. In our last example we walked through how to find the antiderivative just kind of logically working backwards, but for this one I will let you in on the formula. The integral of a power of x is

        ഽx^ndx =( x^n+1/n+1) + c , n =/ -1

What this is telling us to do is to add one to our exponent, and divide by our exponent plus one. Let’s see if this works.

       =18(x^2+1)/2+1  +c = (18x^3)/3 +c= 6x^3 + C

Great! You will eventually, as you continue to practice, get quick enough to just be able to look and see what the original function was before the derivative was taken. But, the formula is super helpful, especially when you start getting into more tricky numbers.

Now, this next one is pretty simple.

ഽkdx = kx + c          Where c and k are both constants.

Here is an example. Let’s say you want to integrate 9:


Our formula tells us that when you have a constant then all you have to do is bring an x to multiply by our 9. Which would give us = 9x +c.

This makes sense, because when we take the derivative of anything times x, the x just goes away, and we are left with a constant.

So, let’s work through a couple different examples.

       ഽ 7x^5 - 4x^-3 + 9 dx

Because, we are adding and subtracting all of our terms we can integrate each term individually.

On our first two terms we can apply the power of x integral formula, and on our nine we can apply our integral of a constant formula.

=((7x^5+1)/5+1) ((-4x^-3 +1)/-3+1) +9x +c
=7/6x^6 +2x-2 +9x +c

Alright, now let’s take a look at another example.

      ഽ7(5throot(x^2)) + 4/x^3 + 1/4sqrt(x) dx

So, again, we can integrate these terms individually, and apply our power of x integral formula. To do that I’m going to rewrite these terms.

         ഽ7x^2/5 + 4x^-3 + 1/4x^-½ dx

Now, I can apply my power of x integral formula.

=(7x^2/5+1 )/(2/5 +1) + (4x^-3 +1)/(-3+1) + (1/4 x^-½ +1)/(-½ + 1) +c
=57/5 - 2x^-2 + ½ sqrt(x) +c

As you practice, you will begin to spot these antiderivatives much easier.

I hope that this video was helpful to you.

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See you guys next time!



by Mometrix Test Preparation | Last Updated: April 15, 2020