����app

Earlier in this course, we learned a rule for finding the derivative of the natural log of x, and we know that this derivative was one over x. Now we also know that we can reverse the process for finding a derivative to find an integral. So, putting those two things together allows us to come up with a new rule for integration that we're going to take a look at here. So let's go ahead and get started. Now, as I just mentioned, the derivative of the natural log of x is one over x, so we can simply reverse this rule to now find the integral of that one over x. So reversing this process here as we integrate one over x, we get back to that natural log.

So the integral of one over x dx is going to be equal to the natural log of the absolute value of x, and then plus our constant of integration c. Now, these absolute value bars are just to account for the fact that we need our antiderivative to be defined for all the same values that our original function was. So this just accounts for any negative values that may occur so that the antiderivative is still defined. Now, another thing that I want to mention here is that you may see this integral written slightly differently because we can rewrite that one over x as x to the power of negative one. So this means the same exact thing.

Remember that having a negative exponent just means that that goes on the bottom of a fraction. So let's go ahead and apply this rule to a quick example here as we integrate five over x dx. Now, five is just a constant, so I can use my constant multiple rule to pull that five out front, making this five times the integral of one over x dx. Now, this is the exact integral that we see up here in our rule, so this is ultimately going to be equal to five times the natural log of the absolute value of x and then plus that constant of integration c. And that's all there is to it.

Now let's take a look at another example here with a slightly more complicated function, as we integrate one over x squared plus three over x dx. Now, depending on how comfortable you are with integrals at this point, you may be able to do this all in one go. But I'm going to go ahead and break this down fully just so that you can see everything that I'm doing along the way. I'm going to go ahead and break this out into two separate integrals using my sum rule. So this becomes the integral of one over x squared dx plus the integral of three over x dx.

Now I can focus on each of these integrals separately. So looking at this first one here, one over x squared dx, because we just learned our rule for finding the integral of one over x, you may be tempted here to think that this will also result in a natural log. But we can't forget all of the rules for integration that we've already learned, and it may be helpful here to rewrite this one over x squared as x to the power of negative two. So we can just use the power rule here. Remember that we can use the power rule to find the integral of x to a negative power so long as that's not x to the power of negative one, because we now know that that will result in a natural log.

But this is x to the power of negative two, so we can go ahead and use our general power rule here, adding one to that exponent and then dividing by that new exponent to make this negative x to the power of negative one. Then, taking a look at our second integral here, three over x dx, three is just a constant. So using my constant multiple rule, I can pull that three out front, making this three times the integral of one over x dx. Now, this is the exact rule that we saw up here. So this will result in three times the natural log of the absolute value of x and then plus my constant of integration c.

Now I can go ahead and rewrite this x to the power of negative one as simply one over x. So this gives me my final answer here, negative one over x plus three times the natural log of the absolute value of x plus c. Now we're going to continue getting practice taking more integrals coming up in the next couple of videos. I'll see you there.

In this problem, we're asked to find the indefinite integral of \( x^3 \times e^{x - 4x^2} \) divided by \( x^3 \) dx. Upon first glance, this might look like a rather complicated integral, but we can actually break it down here. We can see that we have two terms in our numerator and only one term in that denominator so we can effectively split this out into two separate fractions. So let's go ahead and do that here to see if this will help us. Now this is then going to be equal to the integral of \( \frac{x^3 \times e^x}{x^3} \) and then minus \( \frac{4x^2}{x^3} \) dx.

All I've done is split this into two separate fractions. Now I can see here that some stuff is going to cancel out. This \( x^3 \) is going to cancel with that \( x^3 \) and then in this second fraction here, this \( x^2 \) will cancel out two out of those \( x^3 \) making that just \( x \) on the bottom. So that makes this then the integral of \( e^x - \frac{4}{x} \) dx. Now this is a much simpler looking integral that we know how to find.

So this is going to be equal to \( e^x \) because that's what the antiderivative of \( e^x \) is, and then looking at this second term here, this is going to end up being minus four times the natural logarithm of the absolute value of \( x \) based on the rule that we just learned. Then, of course, we want to add on that constant of integration \( c \), and this gives us our final answer here: \( e^x - 4 \times \ln(|x|) + c \). Let us know if you have any questions, and let's keep practicing.