# Video: Indefinite Integrals: Exponential and Reciprocal Functions

In this video, we will learn how to find the indefinite integral of exponential and reciprocal functions (1/𝑥).

14:52

### Video Transcript

In this video, we’re going to learn how to find indefinite integrals of exponential and reciprocal functions. We’ll begin by recalling the first part of the fundamental theorem of calculus before looking at how this helps us to integrate exponential functions of the form 𝑒 to the power of 𝑥 and reciprocal functions of the form 𝑎 over 𝑥.

We begin by stating the first part of the fundamental theorem of calculus. We let 𝑓 be a continuous real-valued function defined on some closed interval 𝑎 to 𝑏. We then let capital 𝐹 be the function defined for all 𝑥 in this closed interval by capital 𝐹 of 𝑥 equals the integral evaluated between 𝑎 and 𝑥 of 𝑓 of 𝑡 with respect to 𝑡. Then, capital 𝐹 is uniformly continuous on the closed interval 𝑎 to 𝑏 and differentiable on the open interval 𝑎 to 𝑏 such that capital 𝐹 prime of 𝑥 is equal to 𝑓 of 𝑥 for all 𝑥 in the open interval 𝑎 to 𝑏. In other words, capital 𝐹 is the antiderivative of a function 𝑓, the function whose derivative is equal to the original function.

And essentially all this tells us is that integration is the reverse process to differentiation. So, let’s begin by looking at the function 𝑓 of 𝑥 equals 𝑒 to the power of 𝑥. Well, we know that the derivative of 𝑒 to the power of 𝑥 is simply 𝑒 to the power of 𝑥. So, the antiderivative of 𝑒 to the power of 𝑥 is also 𝑒 to the power of 𝑥. And we can, therefore, say that the indefinite integral of 𝑒 to the power of 𝑥 evaluated with respect to 𝑥 two is 𝑒 to the power of 𝑥. And of course, since we’re working with an indefinite integral, we add that constant of integration. Let’s call that 𝑐.

But what about the integration of a function 𝑎 to the power of 𝑥 for a real constant 𝑎? Well, once again, we think about derivatives. And we recall the fact that the derivative of 𝑎 to the power of 𝑥 is 𝑎 to the power of 𝑥 times the natural log of 𝑎. We can therefore say that the indefinite integral of 𝑎 to the power of 𝑥 times the natural log of 𝑎 is equal to 𝑎 to the power of 𝑥, since that’s its antiderivative. Once again, we’re working with an indefinite integral, so we add a constant of integration 𝑐.

But this isn’t quite what we’re after. We were wanting to evaluate the integral of 𝑎 of the power of 𝑥. So, we take this constant, the natural log of 𝑎, outside of our integral. And then, we divide both sides by the natural log of 𝑎. And we’ve obtained the integral of 𝑎 to the power of 𝑥 to be 𝑎 to the power of 𝑥 over the natural log of 𝑎 plus 𝐶. Notice, too, that I’ve written this constant as capital 𝐶. That’s to demonstrate that we’ve divided our original constant by another constant, thereby changing this number. Now, it’s useful to know where these results come from, but generally it’s fine to state them. We’re now going to look at some examples demonstrating the integration of these exponential functions.

Determine the indefinite integral of eight 𝑒 to the three 𝑥 minus 𝑒 to the two 𝑥 plus nine over seven 𝑒 to the power of 𝑥 with respect to 𝑥.

Now, this question might look really nasty, and you might be starting to consider how a substitution might help. However, it’s important to notice that we can simplify the integrand by simply dividing each part of the numerator by seven 𝑒 to the power of 𝑥, remembering that we can then simply subtract the exponents. When we do, we’re left with the indefinite integral of eight-sevenths 𝑒 to the two 𝑥 minus one-seventh 𝑒 to the 𝑥 plus nine-sevenths 𝑒 to the negative 𝑥.

Next, we recall that we can separate this integral. The integral of the sum of a number of functions is equal to the sum of the integrals of each function. And we can also take any constant factors outside of the integral and deal with them later. So, we’re looking at eight-sevenths of the integral of 𝑒 to the two 𝑥 with respect to 𝑥 minus one-seventh of the integral of 𝑒 to the 𝑥 with respect to 𝑥 plus nine-sevenths of the integral of 𝑒 to the power of negative 𝑥 with respect to 𝑥. Okay, So what next?

Well, we know that the indefinite integral of 𝑒 to the power of 𝑥 is 𝑒 to the power of 𝑥 plus 𝑐. But what about the integral of 𝑒 of the power of two 𝑥? You might be wanting to make a prediction as to what you think this will give. And there is a standard result that we can quote. But let’s look at this derivation using integration by substitution. We’re going to let 𝑢 be equal to two 𝑥. So, d𝑢 by d𝑥 is equal to two. Now, we know that d𝑢 by d𝑥 is absolutely not a fraction, but we can treat it a little like one for the purposes of integration by substitution. And we see that a half d𝑢 is equal to d𝑥.

And then, we see that we can simply replace two 𝑥 with 𝑢 and d𝑥 with a half d𝑢 and then take out this factor of one-half. And all we need to do now is integrate 𝑒 to the power of 𝑢 with respect to 𝑢. Well, that’s simply a half times 𝑒 to the power of 𝑢. But of course, since 𝑢 is equal to two 𝑥, we can say that the indefinite integral of 𝑒 to the power of two 𝑥 is a half times 𝑒 to the power of two 𝑥 plus a constant of integration. Let’s call that 𝐴. And this is great because it provides us with the general result for the integral of 𝑒 the power of 𝑎𝑥 for real constants 𝑎. It’s one over 𝑎 times 𝑒 to the power of 𝑎𝑥.

We can use this result. And we see that the integral of 𝑒 to the power of negative 𝑥 is one over negative one times 𝑒 to the power of negative 𝑥 plus 𝑐. So, popping this altogether, we see that our indefinite integral is eight-sevenths times a half 𝑒 to the two 𝑥 plus 𝐴 minus a seventh times 𝑒 to the 𝑥 plus 𝐵 plus nine-sevenths times one over negative one times 𝑒 to the power of negative 𝑥 plus 𝐶. Distributing our parentheses and combining all our constants, and we’re left with the solution as being four-sevenths 𝑒 to the power of two 𝑥 minus 𝑒 to power of 𝑥 over seven minus nine-sevenths 𝑒 to the power of negative 𝑥 plus 𝐷.

This example demonstrated really nicely that whilst we can obtain the integral of 𝑒 to the power of 𝑎𝑥 from the general result for 𝑒 of the power of 𝑥 and a clever substitution. It’s much more sensible to use the result given. That is the integral of 𝑒 to the power of 𝑎𝑥 with respect to 𝑥 is one over 𝑎 times 𝑒 to the power of 𝑎𝑥 plus 𝑐. We can even generalize further and say that the integral of 𝑒 to the power of 𝑎𝑥 plus 𝑏 is 𝑒 the power of 𝑎𝑥 plus 𝑏 over 𝑎.

In our next example, we’re going to consider how a substitution can help us obtain the result for the integral of 𝑎 to the power of 𝑏𝑥, where both 𝑎 and 𝑏 are real constants.

Determine the integral of two to the power of nine 𝑥 with respect to 𝑥.

Let’s begin by quoting what we do know about the integral of 𝑎 to the power of 𝑥. It’s 𝑎 to the power of 𝑥 divided by the natural log of 𝑎. Our integrand is slightly different though; it’s a constant to the power of another constant times 𝑥. So, we’re going to use the process of introducing something new, a new letter. We let 𝑢 be equal to nine 𝑥, and of course this is known as integration by substitution. We obtain the derivative of d𝑢 with respect to 𝑥 to be equal to nine. Now, remember, d𝑢 by d𝑥 is absolutely not a fraction, but we can treat it a little like one for the purposes of integration by substitution. And we see that we can say that a ninth d𝑢 equals d𝑥.

We replace 𝑢 with nine 𝑥 and d𝑥 with a ninth d𝑢. And then, we take out this constant factor of a ninth. And we see that our integral is now a ninth of the integral of two to the power of 𝑢 with respect to 𝑢. Well, the integral of two to the power of 𝑢 is two to the power of 𝑢 over the natural log of two. And then, of course, we can use the definition of our substitution and replace 𝑢 with nine 𝑥. And we’ve found the integral of two to the power of nine 𝑥 with respect to 𝑥. It’s two to the power of nine 𝑥 over nine times the natural log of two plus this constant of integration 𝐶, which I’ve made a capital 𝐶 to show that it’s different from the value we had before.

This example gives us a general result for the integration of 𝑎 to the power of 𝑏𝑥 for real constants 𝑎 and 𝑏. It’s 𝑎 to the power of 𝑏𝑥 over 𝑏 times the natural log of 𝑎 plus that constant of integration 𝐶. We’re now going to consider some reciprocals. That is, functions of the form 𝑎 over 𝑥. Now, you might initially think that this is quite straightforward. We can rewrite this function as 𝑎 times 𝑥 to the power of negative one and go from there. But let’s see what happens when we use the power rule for integration.

We add one to the power. That gives us 𝑥 to the power of zero. And then, we divide by zero. We can’t do that though. It gives us a number which is undefined. So, what else can we do? Well, once again, we’re going to consider a derivative. We recall the derivative of the natural log of 𝑥 as being one over 𝑥. And this leads us really nicely into the result that the integral of the one over 𝑥 is the natural log of 𝑥.

But actually, we do need to redefine this a little. The natural log of 𝑥 can only take real positive values of 𝑥 is greater than zero. So, we can instead say that the integral of one over 𝑥 for non-zero values of 𝑥 is the natural log of the absolute value of 𝑥. And of course, we have that constant of integration 𝑐. Generalizing for 𝑎 over 𝑥, we find that the integral of 𝑎 over 𝑥 is 𝑎 times the natural log of the absolute value of 𝑥 plus 𝑐, of course.

Determine the indefinite integral of negative two over seven 𝑥 d𝑥.

We’re going to begin by removing the constant factor from this expression. And that gives us negative two-sevenths times the integral of one over 𝑥 d𝑥. We then quote the general result for the integral of one over 𝑥. It’s the natural log of the absolute value of 𝑥. And so, we obtain our integral to be negative two-sevenths times the natural log of the absolute value of 𝑥 plus 𝑐. Finally, we distribute our parentheses. And we find that the solution to this question is negative two-sevenths times the natural log of the absolute value of 𝑥 plus capital 𝐶.

And notice here I’ve written capital 𝐶 to demonstrate that the original constant has been multiplied by negative two-sevenths, thereby changing it. It’s useful to remember that we can actually check our solution by performing the reverse process, by differentiating. The derivative of the natural log of 𝑥 is, of course, one over 𝑥. So, the derivative of negative two-sevenths times the natural log of 𝑥 is negative two-sevenths times one over 𝑥. And the derivative of a constant 𝐶 is zero. Multiplying, and we end up with our derivative to be negative two over seven 𝑥, as required.

Find, if possible, an antiderivative capital 𝐹 of 𝑓 of 𝑥 equals one over two 𝑥 minus one that satisfies the conditions capital 𝐹 of zero is one and capital 𝐹 of one is negative one.

Another way of thinking about the antiderivative is evaluating the indefinite integral. So, what we’re actually going to do is integrate one over two 𝑥 minus one with respect to 𝑥 and consider the conditions on it in a moment. To integrate one over two 𝑥 minus one, we’re going to use a substitution. We’re going to let 𝑢 be equal to two 𝑥 minus one, which means that d𝑢 by d𝑥 equals two. Now, we do know that d𝑢 by d𝑥 is not a fraction, but we do treat it a little like one for the purposes of integration by substitution and say that a half d𝑢 equals d𝑥.

Let’s perform some substitutions. We replace two 𝑥 minus one with 𝑢 and d𝑥 with a half d𝑢. And we can, of course, take out the constant factor of one-half, and we have half times the integral of one over 𝑢. Well, remember, the integral of one over 𝑥 with respect to 𝑥 is the natural log of the absolute value of 𝑥. So, we can say that the integral of one over two 𝑥 minus one d𝑥 is a half times the natural log of 𝑢 plus 𝑐. We replace 𝑢 with two 𝑥 minus one, and we’ve obtained our antiderivative. Distributing the parentheses, we see that the antiderivative 𝑓 is equal to a half times the natural log of two 𝑥 minus one plus capital 𝐶.

We’re now going to consider conditions. But we also recall the fact that two 𝑥 minus one cannot be equal to zero. So, we’re actually going to have a piecewise function for our antiderivative. The first part of the function we’re interested in is when two 𝑥 minus one is greater than zero. Solving for 𝑥, and we obtain 𝑥 to be greater than one-half. The second is when one minus two 𝑥 is greater than zero. Solving for 𝑥 this time, and we obtain 𝑥 to be less than one-half. Let’s clear some space and formalize this.

We currently have that our antiderivative, capital 𝐹 of 𝑥, is equal to a half times the natural log of one minus two 𝑥 plus some constant when 𝑥 is less than one-half, and a half times the natural log of two 𝑥 minus one plus some constant when 𝑥 is greater than one half. We’re now going to use our conditions on the antiderivative to evaluate 𝑐. The first is capital 𝐹 of zero equals one. So, when 𝑥 is equal to zero, 𝐹 is equal to one.

Since 𝑥 is less than one-half, we’re going to use the first part of our function. We substitute 𝑥 equals zero and capital 𝐹 equals one in. And we get one equals a half times the natural log of one minus two times zero. Well, the natural log of one minus two times zero is the natural log of one, which is of course zero. So, we obtain 𝑐 here to be one.

The second condition we have is that when 𝑥 is equal to one, capital 𝐹 equals negative one. This time 𝑥 is greater than one-half, so we’re going to use the second part of the function. We see that negative one is equal to a half times the natural log of two times one minus one plus 𝑐. Once again, the natural log of two times one minus one is zero. So, we see that 𝑐 in the second part is equal to negative one. And so, the antiderivative capital 𝐹 does indeed exist. Such that capital 𝐹 of 𝑥 is equal to a half times the natural log of one minus two 𝑥 plus one for 𝑥 is less than one-half and a half times the natural log of two 𝑥 minus one minus one for 𝑥 is greater than one-half.

In this video, we’ve seen that we can use the first part of the fundamental theorem of calculus to integrate exponential reciprocal functions. We saw that the integral of 𝑒 the power of 𝑎𝑥 is equal to one over 𝑎 times 𝑒 to the power of 𝑎𝑥. The integral of 𝑎 to the power of 𝑏𝑥 is 𝑎 to the power of the 𝑏𝑥 over 𝑏 times the natural log of 𝑎. And the integral of 𝑎 over 𝑥 with respect to 𝑥 is 𝑎 times the natural log of the absolute value of 𝑥, given that 𝑥 is not equal to zero. And of course, since we’ve been working with indefinite integrals, we need to assume that we’ve got this constant of integration 𝑐 each time.