Lesson Video: Indefinite Integrals: Exponential and Reciprocal Functions | Nagwa Lesson Video: Indefinite Integrals: Exponential and Reciprocal Functions | Nagwa

# Lesson Video: Indefinite Integrals: Exponential and Reciprocal Functions Mathematics

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

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### Video Transcript

Indefinite Integrals: Exponential and Reciprocal Functions

In this video, we will learn how to find the indefinite integral of exponential and reciprocal functions. Weβll do this by recalling two of our derivative results involving exponential and reciprocal functions. We can then use these derivative results to determine antiderivatives of exponential and reciprocal functions. In particular, weβll be able to find the indefinite integral of exponential functions of the form π to the power of ππ₯ and reciprocal functions of the form π over π₯. To do this, we recall to determine the indefinite integral of a function, we want to determine an antiderivative of the function. And one way of doing this is to reverse derivative results. For example, we recall if π of π₯ is the exponential function π to the power of π₯, then we know π prime of π₯ is also π to the power of π₯.

This tells us two things. First, it tells us exactly what it says. We can differentiate π to the power of π₯ to just get π to the power of π₯. However, this also tells us another result: π to the power of π₯ is an antiderivative of π to the power of π₯. Therefore, π to the power of π₯ plus πΆ is the most general antiderivative of this exponential function. In other words, the indefinite integral of π to the power of π₯ with respect to π₯ is π to the power of π₯ plus the constant of integration πΆ. At this point, we could move on to reciprocal functions. However, thereβs two different forms of exponential functions we want to cover first.

Letβs consider exponential functions of the form π to the power of π times π₯, where π is a real constant. And we know how to differentiate functions like this. We recall π prime of π₯ will be equal to π times π to the power of ππ₯. We just multiplied by the coefficient of π₯. And just as we did above, we could turn this into an integral result. The integral of ππ to the power of ππ₯ with respect to π₯ is π to the power of ππ₯ plus πΆ. However, itβs more common to just see functions of the form π to the power of ππ₯. So, we want to remove this value of π. We do this by multiplying our original function by one over π, where we assume our value of π is nonzero. Then, we see that π prime of π₯ is one over π times π times π to the power of ππ₯, which is just π to the power of ππ₯.

Therefore, weβve shown one over π times π to the power of ππ₯ is an antiderivative of π to the power of ππ₯, provided π is nonzero, which also means that weβve shown the indefinite integral of π to the power of ππ₯ with respect to π₯ is one over π times π to the power of ππ₯ plus πΆ, provided π is nonzero.

Thereβs one more type of exponential function we want to look at before we move on to reciprocal functions. So far, all of our exponential functions have had a base of π; however, we can consider what happens if we have a general base. We want to consider the exponential function π of π₯ is equal to π to the power of π₯, where our base needs to be positive and not equal to one. We can recall π prime of π₯ will be equal to π to the power of π₯ multiplied by the natural logarithm of π. Just as we did above, we can write this into an integral result. Weβve shown that π to the power of π₯ is an antiderivative of π to the power of π₯ times the natural logarithm of π.

Therefore, the integral of π to the power of π₯ times the natural logarithm of π with respect to π₯ is equal to π to the power of π₯ plus the constant of integration πΆ, provided π is positive and not equal to one. And since π is not equal to one, we can divide through by the natural logarithm of π. This gives us the indefinite integral of π to the power of π₯ with respect to π₯ is equal to π to the power of π₯ divided by the natural logarithm of π plus the constant of integration πΆ. Well, itβs worth noting we can write πΆ in both of these cases because πΆ is just a constant. Dividing through by the natural logarithm of π still leaves this as a constant, so we can just call this πΆ in both cases. Letβs now move on to the final example of reciprocal functions.

To determine the indefinite integral of a reciprocal function, we want a function which differentiates to give us the reciprocal function. And we can do this by recalling if π of π₯ is the natural logarithm of the absolute value of π₯, then π prime of π₯ will be equal to one over π₯. Therefore, the natural logarithm of the absolute value of π₯ is an antiderivative of one over π₯, which tells us the indefinite integral of one over π₯ with respect to π₯ is the natural logarithm of the absolute value of π₯ plus πΆ. And we can generalize this result slightly by multiplying our original function through by π.

If π of π₯ is π times the natural logarithm of the absolute value of π₯, then π prime of π₯ is π over π₯, which then tells us for any real constant π the integral of π over π₯ with respect to π₯ is π times the natural logarithm of the absolute value of π₯ plus πΆ. Letβs now look at some examples where we use these results to evaluate the integral of more complicated functions.

Determine the indefinite integral of eight times π to the power of three π₯ minus π to the power of two π₯ plus nine divided by seven π to the power of π₯ with respect to π₯.

In this question, weβre asked to evaluate the indefinite integral of an exponential expression. And at first, it might look very difficult to evaluate this integral. For example, we might be considering a π’-substitution. However, we should always check if we can simplify our integrand. And we can do this by dividing every term in the numerator by the denominator, where we remember for each term in the numerator, we just need to subtract the exponents of π. This gives us the indefinite integral of eight over seven times π to the power of two π₯ minus one-seventh π to the power of π₯ plus nine over seven times π to the power of negative π₯ with respect to π₯.

Now, we can see we know how to integrate each term in the integrand separately. So, weβre going to split this into three separate integrals. And we can do this because we recall the integral of the sum of functions is equal to the sum of their individual integrals. This gives us the following. We can then simplify each term separately. Weβll take the constant factor outside of each integral. This then gives us eight-sevenths times the indefinite integral of π to the power of two π₯ with respect to π₯ minus one-seventh the indefinite integral of π to the power of π₯ with respect to π₯ plus nine-sevenths the integral of π to the power of negative π₯ with respect to π₯.

Weβre now ready to evaluate each of these indefinite integral separately. We can do this by recalling for any real constant π not equal to zero, the integral of π to the power of ππ₯ with respect to π₯ is one over π times π to the power of ππ₯ plus the constant of integration πΆ. Weβll use this to evaluate each integral separately. Letβs start with the first integral, the integral of π to the power of two π₯ with respect to π₯. Our value of π is two. So, we substitute π is two into our integral rule. We get one-half times π to the power of two π₯. And we need to multiply this by eight over seven. And itβs worth noting since weβre evaluating the sum of multiple indefinite integrals, weβll get a constant of integration for each of these integrals. We can combine all of these constants into one constant at the end of our expression. So, we donβt need to worry about adding πΆ until the end.

Letβs now evaluate our second integral. We know π to the power of π₯ is the same as π to the power of one times π₯. So, our value of π will be one. Of course, this is somewhat not necessary since we know the integral of π to the power of π₯ with respect to π₯ is just π to the power of π₯ with respect to π₯. However, if we did substitute π is equal to one, we would get one over one times π to the power of one π₯, which is just π to the power of π₯. So, evaluating our second integral and multiplying by negative one-seventh, we get negative one-seventh π to the power of π₯.

Finally, letβs evaluate the third integral. Our value of π is negative one. Substituting π is negative one into our integral result gives us one over negative one times π to the power of negative π₯. We need to multiply this by nine over seven. And remember, we also need to add a constant of integration at the end of this expression. We can then simplify this slightly. One over negative one is just equal to negative one. So, instead of adding this term, we can subtract this term. And we can also notice in the first term, we have a shared factor of two in the numerator and denominator. We can cancel this to get a factor of four over seven.

This then gives us our final answer. The indefinite integral of π to the power of three π₯ minus π to the power two π₯ plus nine over seven π to the π₯ with respect to π₯ is equal to four over seven times π to the power of two π₯ minus π to the power of π₯ over seven minus nine-sevenths π to the power of negative π₯ plus the constant of integration πΆ.

In our previous example, we were able to rewrite our integrand into a form where we could integrate each term by using our exponential integral rules. And although it was more difficult, we also noted we couldβve tried using the π’-substitution. And weβll have a similar fort process in the next example. We can either attempt to do this by using a π’-substitution or rewriting our integrand.

Determine the indefinite integral of two to the power of nine π₯ with respect to π₯.

In this question, weβre asked to evaluate the integral of an exponential function. And we can note this looks very similar to one of our integral results. For any positive constant π not equal to one, the integral of π to the power of π₯ with respect to π₯ is equal to π to the power of π₯ divided by the natural logarithm of π plus the constant of integration πΆ. However, in this case, our exponent is not just equal to π₯; we have nine π₯. So, we canβt just directly apply this integral result. And there are two ways we can try and fix this. We could try using the π’-substitution. π’ is nine π₯. This would then allow us to rewrite our integral in this form. However, thereβs actually a simpler method.

We can just apply our laws of exponents. Two to the ninth power all raised to the power of π₯ will be equal to two to the power of nine times π₯. So, we can rewrite our integral as the integral of two to the ninth power all raised to the power of π₯ with respect to π₯. Then, our value of π is two to the ninth power. We could now evaluate two to the ninth power as 512. However, weβre just going to substitute π is equal to two to the power of nine into our integral result. This then gives us two to the ninth power all raised to the power of π₯ all divided by the natural logarithm of two to the ninth power plus the constant of integration πΆ. And this can help us see why itβs useful to leave this as two to the ninth power rather than evaluate.

We can simplify the numerator by using our laws of exponents, and we can simplify our denominator by using the power rule for logarithms. First, in the numerator, two raised to the power of nine all raised to the power of π₯ will be two to the power of nine times π₯. Next, in the denominator, the power rule for logarithms tells us the natural logarithm of two to the ninth power will be nine times the natural logarithm of two, which then gives us our final answer. The integral of two to the power of nine π₯ with respect to π₯ is two to the power of nine π₯ divided by nine times the natural logarithm of two plus the constant of integration πΆ.

Itβs worth noting we can use the method we used in this question to prove a general integral result. We can apply this method to the indefinite integral of π to the power of ππ₯ with respect to π₯. We rewrite our integrand as π to the power of π all raised to the power of π₯ and then use our integral result. Then, we rewrite the numerator by using our laws of exponents and the denominator by using the power rule for logarithms. We get π to the power of ππ₯ divided by π times the natural logarithm of π plus the constant of integration πΆ.

Letβs now see an example of integrating a reciprocal function.

Determine the indefinite integral of negative two divided by seven π₯ with respect to π₯.

In this question, weβre asked to evaluate the indefinite integral of a reciprocal function. And we can do this directly by recalling one of our integral results. For any real constant π, the indefinite integral of π over π₯ with respect to π₯ is equal to π times the natural logarithm of the absolute value of π₯ plus the constant of integration πΆ. We can just note that our value of π will be negative two over seven; however, sometimes it can be difficult to see this. So, instead, weβll just take the constant factor of negative two over seven outside of our integral. This leaves us with negative two over seven times the indefinite integral of the reciprocal function, one over π₯ with respect to π₯.

And we know the integral of the reciprocal function is the natural logarithm of the absolute value of π₯. We can just recall this, or we can set our value of π equal to one into our integral result. Using either method, weβve shown the indefinite integral of negative two over seven π₯ with respect to π₯ is negative two-sevenths times the natural logarithm of the absolute value of π₯ plus πΆ. And itβs worth noting we can always check our answer by using differentiation. Remember, when weβre finding the indefinite integral of a function, weβre finding its most general antiderivative. This means when we differentiate our function with respect to π₯, we should end up with our integrand.

So, letβs evaluate the derivative of negative two-sevenths times the natural logarithm of the absolute value of π₯ plus πΆ with respect to π₯. We can start by recalling the derivative of the natural logarithm of the absolute value of π₯ with respect to π₯ is one over π₯. So, when we differentiate the first term with respect to π₯, we get negative two over seven multiplied by one over π₯. The second term is a constant, so its rate of change with respect to π₯ is zero. This just leaves us with negative two-sevenths times one over π₯, which we can simplify is negative two over seven π₯. This is our integrand, so this confirms that our answer is correct. Therefore, the indefinite integral of negative two over seven π₯ with respect to π₯ is negative two over seven times the natural logarithm of the absolute value of π₯ plus πΆ.

Letβs now see an example where we need to use boundary conditions to determine a specific antiderivative of a given reciprocal function.

Find, if possible, an antiderivative capital πΉ of lowercase π of π₯ is equal to one divided by two π₯ minus one that satisfies the conditions capital πΉ evaluated at zero is one and capital πΉ evaluated at one is negative one.

In this question, weβre asked to find an antiderivative of a given reciprocal function that satisfies two conditions; these are called boundary conditions. We can do this by recalling we can determine antiderivatives of functions by using indefinite integration. In particular, the indefinite integral of one divided by two π₯ minus one with respect to π₯ will give us the most general antiderivative of this function. We know how to integrate the reciprocal function. However, in this case, we can see our denominator is a linear function. So, weβre going to use a π’-substitution.

Weβre going to let π’ be equal to the denominator, two π₯ minus one. Now, remember to integrate by substitution, we need to find an equation involving the differentials. And we can do this by differentiating both sides of our substitution with respect to π₯. Since two π₯ minus one is a linear function, its derivative with respect to π₯ will be the coefficient of π₯, which is two. Now, although dπ’ by dπ₯ is not a fraction, we can treat it a little bit like a fraction when weβre using integration by substitution. This will allow us to find an equation involving the differentials. This then gives us that a half dπ’ is equal to dπ₯.

Now, we can use integration by substitution. This will allow us to rewrite our integral. We rewrite the denominator of our integrand as π’, and we replace dπ₯ with a half dπ’. We get the indefinite integral of one over π’ times a half with respect to π’. And we can simplify our integral slightly by taking the constant factor of a half outside of our integral. This gives us a half times the indefinite integral of one over π’ with respect to π’. We can now evaluate this integral by recalling the integral of one over π₯ with respect to π₯ is equal to the natural logarithm of the absolute value of π₯ plus the constant of integration πΆ. In our case, our variable is π’. So, we get one-half times the natural logarithm of the absolute value of π’ plus the constant of integration πΆ.

Well, itβs worth noting we donβt need to multiply our constant by one-half since one-half times a constant is still a constant, so weβll just call this πΆ. But remember, our original integrand is in terms of π₯ and our original function lowercase π of π₯ is also in terms of π₯. So, we should rewrite our answer to be in terms of the variable π₯. We can do this by using our π’-substitution. We get a half times the natural logarithm of the absolute value of two π₯ minus one plus πΆ. And at this point, we might be tempted to call this our function capital πΉ and then start substituting π₯ is equal to zero and π₯ is equal to one and solving for πΆ. However, while this method would usually work, it wonβt work in this case because weβre actually using shorthand notation.

Weβre taking the natural logarithm of the absolute value of two π₯ minus one. This means that this is a piecewise function. And this makes a lot of sense because to differentiate the natural logarithm of the absolute value of π₯, we do this in two parts. The derivative of the natural logarithm of π₯ when π₯ is greater than zero is one over π₯, and the derivative of the natural logarithm of negative π₯ is one over π₯ when π₯ is less than zero. In the same way, our integral result will actually result in a piecewise function, depending on whether our argument is positive or negative.

So, weβre going to need to write this as a piecewise function. Weβll do this by noticing when π₯ is greater than one-half, two π₯ minus one will be positive, and when π₯ is less than one-half, two π₯ minus one will be negative. So, if our value of π₯ is greater than one-half, then the absolute value of two π₯ minus one is just equal to two π₯ minus one. And if π₯ is less than one-half, then two π₯ minus one is negative. So, the absolute value of two π₯ minus one will be negative one times two π₯ minus one, which is one minus two π₯.

This allows us to write our piecewise function capital πΉ of π₯. Itβs equal to a half times the natural logarithm of two π₯ minus one plus the constant of integration weβve called πΆ sub one when π₯ is greater than one-half. And itβs equal to one-half times the natural logarithm of one minus two π₯ plus the constant of integration πΆ sub two if π₯ is less than one-half. And the important thing to note here is our constants of integration donβt need to be the same. Because remember, when we differentiate this function, weβll differentiate each subfunction separately over its subdomain. And the derivatives of these constants will always be equal to zero; it doesnβt matter what their values are.

And finally, it may be worth noting π₯ is equal to one-half is not included in either subdomain. So, we donβt need to worry about this if we were to differentiate capital πΉ of π₯. This is simply because one-half is not in the domain of our original function, lowercase π of π₯. We can now determine the values of πΆ sub one and πΆ sub two from the question. Weβll start by substituting π₯ is equal to zero into this function. To substitute π₯ is equal to zero into our piecewise-defined function, we note that zero is less than one-half, so we need to use our second subfunction. We get that capital πΉ evaluated at zero is equal to a half times the natural logarithm of one minus two times zero plus πΆ sub two. Weβre told in the question capital πΉ evaluated at zero is equal to one.

Next, one minus two times zero is equal to one, and the natural logarithm of one is just equal to zero. So, this term just simplifies to give us zero. Therefore, weβve shown that one is equal to πΆ sub two. We can now substitute π₯ is equal to one into our function capital πΉ to determine the value of πΆ sub one. Since one is greater than one-half, this is in the first sub domain of our piecewise-defined function. So, capital πΉ evaluated at one is one-half times the natural logarithm of two times one minus one plus πΆ sub one. Weβre told in the question capital πΉ evaluated at one is negative one, and we can note two times one minus one is one. And the natural logarithm of one is just zero. So, this simplifies to give us that πΆ sub one is equal to negative one.

Now, all we need to do is substitute our values for πΆ sub one and πΆ sub two into our function capital πΉ of π₯. This then gives us the following. And itβs worth noting we can technically write our subfunctions in any order; however, usually we write this so the subdomain on the left is at the top. And since π₯ being less than one-half starts at the left side of our number line, weβll write this at the top of our function. And this then gives us our final answer. Capital πΉ of π₯ will be equal to a half times the natural logarithm of one minus two π₯ plus one if π₯ is less than one-half. And capital πΉ of π₯ will be equal to a half times the natural logarithm of two π₯ minus one minus one if π₯ is greater than one-half.

Letβs now go over the key points of this video. First, we were able to reverse our derivative results involving exponential functions to construct equivalent results about indefinite integrals. For example, we were able to show the indefinite integral of π to the power of π₯ with respect to π₯ is π to the power of π₯ plus πΆ. And for any real constant π not equal to zero, the indefinite integral of π to the power ππ₯ with respect to π₯ is equal to one over π times π to the power of ππ₯ plus πΆ.

And finally, for a positive real constant π not equal to one, the indefinite integral of π to the power of π₯ with respect to π₯ is π to the power of π₯ divided by the natural logarithm of π plus πΆ. Secondly, we were able to reverse another derivative result to show for any real constant π the indefinite integral of π over π₯ with respect to π₯ is π times the natural logarithm of the absolute value of π₯ plus πΆ.