Lesson Explainer: Indefinite Integrals: Exponential and Reciprocal Functions

In this explainer, we will learn how to find the indefinite integral of exponential and reciprocal functions ๏€ผ1๐‘ฅ๏ˆ.

Indefinite integrals of exponential and logarithmic functions have many real-world applications as the functions are used in mathematical models to describe population growth, cell growth, and radioactive decay.

These types of problems can be solved by using the following rules.

Definition: Integrals of Exponential and Reciprocal Functions

๏„ธ๐‘’๐‘ฅ=1๐‘Ž๐‘’+๏Œบ๏—๏Œบ๏—dC(1)

and

๏„ธ๐‘Ž๐‘ฅ๐‘ฅ=๐‘Ž|๐‘ฅ|+,dlnC(2)

for ๐‘Žโˆˆโ„โˆ’{0} to avoid dividing by zero in the first expression.

We can verify these directly using the first part of the fundamental theorem of calculus.

Definition: The Fundamental Theorem of Calculus (FTC)

Let ๐‘“ be a continuous real-valued function defined on [๐‘Ž,๐‘]. Let ๐น be the function defined, for all ๐‘ฅ in [๐‘Ž,๐‘], by ๐น(๐‘ฅ)=๏„ธ๐‘“(๐‘ก)๐‘ก.๏—๏Œบd

Then, ๐น is uniformly continuous on [๐‘Ž,๐‘] and differentiable on (๐‘Ž,๐‘), and ๐นโ€ฒ(๐‘ฅ)=๐‘“(๐‘ฅ), for all ๐‘ฅ in (๐‘Ž,๐‘).

By taking the derivative of the right-hand side for each of the statements above, we can show the result equals the integrand: ddddln๐‘ฅ๏€ผ1๐‘Ž๐‘’๏ˆ=1๐‘Ž(๐‘Ž๐‘’)=๐‘’,๐‘ฅ(๐‘Ž|๐‘ฅ|)=๐‘Ž๐‘ฅ.๏Œบ๏—๏Œบ๏—๏Œบ๏—

Therefore, upon integrating these, we have ๏„ธ๐‘’๐‘ฅ=๏„ธ๐‘ฅ๏€ผ1๐‘Ž๐‘’๏ˆ๐‘ฅ=1๐‘Ž๐‘’+,๏„ธ๐‘Ž๐‘ฅ๐‘ฅ=๏„ธ๐‘ฅ(๐‘Ž|๐‘ฅ|)๐‘ฅ=๐‘Ž|๐‘ฅ|+.๏Œบ๏—๏Œบ๏—๏Œบ๏—ddddCdddlndlnC

Now, what if we want to integrate something like ๐‘Ž๏— (i.e., an exponential expression with an arbitrary base)? The trick is to use the fact that ๐‘’=๐‘กln๏ and apply the laws of logarithms with lnln๐‘Ž=๐‘ฅ๐‘Ž๏—: ๐‘Ž=๐‘’=๐‘’.๏—๏Œบ๏—๏Œบlnln๏‘

We can use the standard rule for exponentials (1) with base ๐‘’ to obtain ๏„ธ๐‘Ž๐‘ฅ=๏„ธ๐‘’๐‘ฅ=1๐‘Ž๐‘’+=1๐‘Ž๐‘Ž+,๏—๏—๏Œบ๏—๏Œบ๏—ddlnClnClnln where ๐‘Žโˆˆโ„โˆ’{1}๏Šฐ to avoid dividing by zero. More generally, we have ๏„ธ๐‘Ž๐‘ฅ=1๐‘๐‘Ž๐‘Ž+,๏Œป๏—๏Œป๏—dlnC for ๐‘Žโˆˆโ„โˆ’{1}๏Šฐ and ๐‘โˆˆโ„โˆ’{0}. We do not need to memorize this result as we can derive it directly from applying the laws of logarithms for each exponential base.

Letโ€™s look at an integral with both exponential and reciprocal functions given by ๐ผ(๐‘ฅ)=๏„ธ๏€ผ92๐‘ฅโˆ’12๐‘’๏ˆ๐‘ฅ.๏Šฑ๏Šฉ๏—d

We know that the integral of the sum or difference of two functions is equal to the sum or difference of the integrals of those functions. In other words, we can separate out the two parts and take constant factors outside the integral. For each part, we also get a constant of integration, which we can combine into a single constant C.

Applying the standard rules for integrating exponential (1) and reciprocal (2) functions as given in the definition, we find ๐ผ(๐‘ฅ)=92๏„ธ1๐‘ฅ๐‘ฅโˆ’12๏„ธ๐‘’๐‘ฅ=92|๐‘ฅ|โˆ’12๏€ผโˆ’13๐‘’๏ˆ+=92|๐‘ฅ|+4๐‘’+.ddlnClnC๏Šฑ๏Šฉ๏—๏Šฑ๏Šฉ๏—๏Šฑ๏Šฉ๏—

The constant of integration can also be determined if there is a boundary condition. Suppose ๐ผ(1)=0 is a boundary condition. We can substitute this in order to determine the constant C as ๐ผ(1)=92|1|+4๐‘’+=0,lnC๏Šฑ๏Šฉ(๏Šง) which after rearranging gives C=โˆ’4๐‘’.๏Šฑ๏Šฉ

Therefore, ๐ผ(๐‘ฅ) can be written as ๐ผ(๐‘ฅ)=92|๐‘ฅ|+4๐‘’โˆ’4๐‘’.ln๏Šฑ๏Šฉ๏—๏Šฑ๏Šฉ

Now letโ€™s look at a few examples to practice and deepen our understanding. The first two examples contain exponential functions of different bases.

Example 1: Finding the Integration of a Function Containing Exponential Functions by Distributing the Division

Determine ๏„ธ8๐‘’โˆ’๐‘’+97๐‘’๐‘ฅ๏Šฉ๏—๏Šจ๏—๏—d.

Answer

In this example, we want to find the indefinite integral of a function containing exponentials with base ๐‘’.

We begin by separating each part of the numerator and dividing individually by ๐‘’๏— in the integrand to give us 8๐‘’โˆ’๐‘’+97๐‘’=87๐‘’๐‘’โˆ’17๐‘’๐‘’+971๐‘’=87๐‘’โˆ’๐‘’7+97๐‘’.๏Šฉ๏—๏Šจ๏—๏—๏Šฉ๏—๏—๏Šจ๏—๏—๏—๏Šจ๏—๏—๏Šฑ๏—

Next, separating the integral and taking any constant factors outside, we can use the standard rule for integrating exponential terms: ๏„ธ๐‘’๐‘ฅ=1๐‘Ž๐‘’+,๐‘Žโˆˆโ„โˆ’{0}.๏Œบ๏—๏Œบ๏—dC

We obtain a constant of integration for these separate parts, but we can combine them into a single constant C. In particular, we have ๏„ธ8๐‘’โˆ’๐‘’+97๐‘’๐‘ฅ=๏„ธ๏€ฝ87๐‘’โˆ’๐‘’7+97๐‘’๏‰๐‘ฅ=87๏„ธ๐‘’๐‘ฅโˆ’17๏„ธ๐‘’๐‘ฅ+97๏„ธ๐‘’๐‘ฅ=87๏€ผ12๐‘’๏ˆโˆ’17(๐‘’)+97(โˆ’๐‘’)+=47๐‘’โˆ’๐‘’7โˆ’97๐‘’+.๏Šฉ๏—๏Šจ๏—๏—๏Šจ๏—๏—๏Šฑ๏—๏Šจ๏—๏—๏Šฑ๏—๏Šจ๏—๏—๏Šฑ๏—๏Šจ๏—๏—๏Šฑ๏—dddddCC

In our first example, we saw how to integrate a number of exponential functions with base ๐‘’. We will now demonstrate how we can use the laws of exponentials and logarithms to integrate an exponential function with a base of 2.

Example 2: Finding the Integration of an Exponential Function with an Integer Base

Determine ๏„ธ2๐‘ฅ๏Šฏ๏—d.

Answer

In this example, we want to find the indefinite integral of a function containing an exponential with base 2.

We first begin by rewriting the integrand in base ๐‘’ as 2=๐‘’=๐‘’,๏Šฏ๏—๏Šจ๏Šฏ๏—๏Šจlnln๏Žจ๏‘ where we have applied the laws of logarithms to obtain the final form. We can now use the standard rule for integrating exponential terms: ๏„ธ๐‘’๐‘ฅ=1๐‘Ž๐‘’+,๐‘Žโˆˆโ„โˆ’{0}.๏Œบ๏—๏Œบ๏—dC

In particular, we find ๏„ธ2๐‘ฅ=๏„ธ๐‘’๐‘ฅ=192๐‘’+=1922+=292+.๏Šฏ๏—๏Šฏ๏—๏Šจ๏Šฏ๏—๏Šจ๏Šฏ๏—๏Šฏ๏—ddlnClnClnClnln

In our next example, we will demonstrate how to apply the rule (2), given in the definition, to integrate reciprocal functions.

Example 3: Finding the Integration of a Reciprocal Function

Determine ๏„ธโˆ’27๐‘ฅ๐‘ฅd.

Answer

In this example, we want to find the indefinite integral of a reciprocal function.

This integral can be performed simply by taking the constant outside the integral and applying the standard rule for the integral reciprocal functions: ๏„ธ๐‘Ž๐‘ฅ๐‘ฅ=๐‘Ž|๐‘ฅ|+,๐‘Žโˆˆโ„.dlnC

In particular, we have ๏„ธโˆ’27๐‘ฅ๐‘ฅ=โˆ’27๏„ธ1๐‘ฅ๐‘ฅ=โˆ’27|๐‘ฅ|+.ddlnC

Now, letโ€™s look at an example where we have to apply the same rules to an integral after distributing the division.

Example 4: Finding the Integration of a Rational Function by Distributing the Division

Determine ๏„ธ(โˆ’2๐‘ฅโˆ’5)๐‘ฅ๐‘ฅ๏Šจd.

Answer

In this example, we want to find the indefinite integral of a rational function. Often, our instinct in these cases might be to apply integration by substitution. However, we notice that if we distribute the parentheses, we can simplify the integrand by dividing each term of the numerator by the ๐‘ฅ in the denominator.

To begin, we distribute the parentheses in the integrand to obtain a quadratic expression in the numerator and then perform the division by ๐‘ฅ: (โˆ’2๐‘ฅโˆ’5)๐‘ฅ=4๐‘ฅ+20๐‘ฅ+25๐‘ฅ=(4๐‘ฅ+20)+25๐‘ฅ.๏Šจ๏Šจ

The first expression in the integrand contains a linear term, which can be integrated using the power rule, and the second expression is a reciprocal, which can be integrated using the standard rule for integrating reciprocals: ๏„ธ๐‘Ž๐‘ฅ๐‘ฅ=๐‘Ž|๐‘ฅ|+,๐‘Žโˆˆโ„.dlnC

In particular, we find ๏„ธ(โˆ’2๐‘ฅโˆ’5)๐‘ฅ๐‘ฅ=๏„ธ๏€ผ(4๐‘ฅ+20)+25๐‘ฅ๏ˆ๐‘ฅ=๏„ธ(4๐‘ฅ+20)๐‘ฅ+25๏„ธ1๐‘ฅ๐‘ฅ=2๐‘ฅ+20๐‘ฅ+25|๐‘ฅ|+.๏Šจ๏ŠจddddlnC

Until this stage, we have applied our rules to find the general solution to integrals involving exponentials and logarithms. We can find a particular solution to these sorts of problems by applying a boundary condition to help us determine the constant of integration.

Example 5: Applying the Fundamental Theorem of Calculus to Integrate a Rational Function with Boundary Conditions

Find, if possible, an antiderivative ๐น of ๐‘“(๐‘ฅ)=12๐‘ฅโˆ’1 that satisfies the conditions ๐น(0)=1 and ๐น(1)=โˆ’1.

Answer

In this example, we want to find the antiderivative of a reciprocal function satisfying particular boundary conditions.

We begin by using the first part of the fundamental theorem of calculus to determine ๐น(๐‘ฅ); it is the indefinite integral of ๐‘“(๐‘ฅ): ๐น(๐‘ฅ)=๏„ธ๐‘“(๐‘ฅ)๐‘ฅ=๏„ธ12๐‘ฅโˆ’1๐‘ฅ.dd

We might recognize the result, or we can use integration by substitution. For the latter, we note that the integrand contains a composite function: ๐‘“(๐‘”(๐‘ฅ))=12๐‘ฅโˆ’1, with ๐‘“(๐‘ฅ)=1๐‘ฅ and ๐‘”(๐‘ฅ)=2๐‘ฅโˆ’1. Thus, we can use the substitution: ๐‘ข=๐‘”(๐‘ฅ)=2๐‘ฅโˆ’1.

The derivative of this with respect to ๐‘ฅ is dd๐‘ข๐‘ฅ=2 or, equivalently, by manipulating the differentials, dd๐‘ฅ=๐‘ข2.

Next, we apply this substitution to the integral to change the variable from ๐‘ฅ to ๐‘ข and integrate the resulting expression using the standard rule for integrating reciprocal functions given by ๏„ธ๐‘Ž๐‘ข๐‘ข=๐‘Ž|๐‘ข|+,๐‘Žโˆˆโ„.dlnC

Thus, we obtain ๐น(๐‘ฅ)=๏„ธ12๐‘ฅโˆ’1๐‘ฅ=๏„ธ1๐‘ข๏€ฝ๐‘ข2๏‰=12๏„ธ1๐‘ข๐‘ข=12|๐‘ข|+.dddlnC

Now, we apply the reverse substitution ๐‘ข=2๐‘ฅโˆ’1 to get everything back in terms of ๐‘ฅ: ๐น(๐‘ฅ)=12|2๐‘ฅโˆ’1|+.lnC

Now, we can use the boundary conditions ๐น(0)=1 and ๐น(1)=โˆ’1. Since our solution to the integral involves a modulus, we recall the definition of |๐‘ฅ|: |๐‘ฅ|=๏ญ๐‘ฅ๐‘ฅโ‰ฅ0,โˆ’๐‘ฅ๐‘ฅ<0.forfor

We also note that the natural log, ln๐‘ฅ, is undefined at ๐‘ฅ=0; thus, for ln|๐‘ฅ|, we have to consider values of ๐‘ฅโ‰ 0: lnlnforlnfor|๐‘ฅ|=๏ฎ๐‘ฅ๐‘ฅ>0,(โˆ’๐‘ฅ)๐‘ฅ<0.

We can also split up the constant of integration C, in each part of the piecewise function, for some ๐‘Žโˆˆโ„ as CCforCfor=๏ฎ๐‘ฅ>๐‘Ž,๐‘ฅ<๐‘Ž.๏Šง๏Šจ

This is essential since the constant will be different, as we shall see, depending on the value of ๐‘ฅ where the function is defined. Therefore, ๐น(๐‘ฅ)=12|2๐‘ฅโˆ’1|+=โŽงโŽจโŽฉ12(2๐‘ฅโˆ’1)+2๐‘ฅโˆ’1>0โ‡”๐‘ฅ>12,12(1โˆ’2๐‘ฅ)+2๐‘ฅโˆ’1<0โ‡”๐‘ฅ<12.lnClnCforlnCfor๏Šง๏Šจ

The boundary condition ๐น(0)=1 can only be applied to the second part with ๐‘ฅ<12 as ๐‘ฅ=0. This allows us to determine C๏Šจ as ๐น(0)=12(1โˆ’2(0))+=1=12(1)+=;lnClnCC๏Šจ๏Šจ๏Šจ hence, C๏Šจ=1.

Similarly, the boundary condition ๐น(1)=โˆ’1 can only be applied to the first part with ๐‘ฅ>12 as ๐‘ฅ=1. This allows us to determine C๏Šง: ๐น(1)=12(2(1)โˆ’1)+=โˆ’1=12(1)+=;lnClnCC๏Šง๏Šง๏Šง hence, C๏Šง=โˆ’1.

Therefore, the antiderivative ๐น of ๐‘“(๐‘ฅ), which satisfies the given boundary conditions, is given by ๐น(๐‘ฅ)=โŽงโŽจโŽฉ12(1โˆ’2๐‘ฅ)+1๐‘ฅ<1212(2๐‘ฅโˆ’1)โˆ’1๐‘ฅ>12.lnforlnfor

In our next example, we will look at how to apply the rules for integrating reciprocal functions involving a root and negative exponents.

Example 6: Finding the Integration of a Function Involving Expanding Squares and Using the Power Rule with Negative Exponents

Determine ๏„ธ๏€ฟ3โˆš๐‘ฅ+79โˆš๐‘ฅ๏‹๐‘ฅ๏Šจd.

Answer

In this example, we want to find the indefinite integral of a function involving โˆš๐‘ฅ and expanding squares.

We begin by distributing the square in the integrand: ๏€ฟ3โˆš๐‘ฅ+79โˆš๐‘ฅ๏‹=9๐‘ฅ+143+4981๐‘ฅ.๏Šจ

Now, we can separate the integrals into two parts: the first expression containing a linear term that can be integrated using the power rule and the second expression a reciprocal term that can be integrated using the standard rule: ๏„ธ๐‘Ž๐‘ฅ๐‘ฅ=๐‘Ž|๐‘ฅ|+,๐‘Žโˆˆโ„.dlnC

In particular, we find ๏„ธ๏€ฟ3โˆš๐‘ฅ+79โˆš๐‘ฅ๏‹๐‘ฅ=๏„ธ๏€ผ9๐‘ฅ+143+4981๐‘ฅ๏ˆ๐‘ฅ=๏„ธ๏€ผ9๐‘ฅ+143๏ˆ๐‘ฅ+4981๏„ธ1๐‘ฅ๐‘ฅ=9๐‘ฅ2+14๐‘ฅ3+4981|๐‘ฅ|+.๏Šจ๏ŠจddddlnC

In our final example, we will apply the integration of exponentials twice in order to determine a function from its second derivative.

Example 7: Finding the Expression of a Function given Its Second Derivative Using Indefinite Integration

Given that ๐‘“โ€ฒโ€ฒ(๐‘ฅ)=โˆ’5๐‘’+2๐‘ฅ๏Šช๏—๏Šซ, find ๐‘“(๐‘ฅ).

Answer

In this example, we want to find the function ๐‘“(๐‘ฅ) from the expression we have for its second derivative. Since integration is the reverse process of differentiation, we can determine ๐‘“(๐‘ฅ) by performing two successive integrals.

We begin first by finding ๐‘“โ€ฒ(๐‘ฅ) by using the first part of the fundamental theorem of calculus. Integrating the expression for the second derivative using the general power rule and the standard rule for exponentials, ๏„ธ๐‘’๐‘ฅ=1๐‘Ž๐‘’+,๐‘Žโˆˆโ„โˆ’{0},๏Œบ๏—๏Œบ๏—dC we obtain ๐‘“โ€ฒ(๐‘ฅ)=๏„ธ๐‘“โ€ฒโ€ฒ(๐‘ฅ)๐‘ฅ=๏„ธ๏€นโˆ’5๐‘’+2๐‘ฅ๏…๐‘ฅ=โˆ’5๏„ธ๐‘’๐‘ฅ+๏„ธ2๐‘ฅ๐‘ฅ=โˆ’5๏€ผ14๐‘’๏ˆ+13๐‘ฅ+=โˆ’54๐‘’+13๐‘ฅ+.ddddCC๏Šช๏—๏Šซ๏Šช๏—๏Šซ๏Šช๏—๏Šฌ๏Šช๏—๏Šฌ

Finally, we determine ๐‘“(๐‘ฅ) by using the fundamental theorem of calculus again and performing another integral with the expression of the first derivative: ๐‘“(๐‘ฅ)=๏„ธ๐‘“โ€ฒ(๐‘ฅ)๐‘ฅ=๏„ธ๏€ผโˆ’54๐‘’+13๐‘ฅ+๏ˆ๐‘ฅ=โˆ’54๏„ธ๐‘’๐‘ฅ+๏„ธ๏€ผ13๐‘ฅ+๏ˆ๐‘ฅ=โˆ’54๏€ผ14๐‘’๏ˆ+๐‘ฅ21+๐‘ฅ+=โˆ’516๐‘’+๐‘ฅ21+๐‘ฅ+.dCddCdCDCD๏Šช๏—๏Šฌ๏Šช๏—๏Šฌ๏Šช๏—๏Šญ๏Šช๏—๏Šญ

Key Points

  • Indefinite integrals involving exponentials and reciprocals can be evaluated using the standard results: ๏„ธ๐‘’๐‘ฅ=1๐‘Ž๐‘’+.๏Œบ๏—๏Œบ๏—dC
  • To find the integral of an exponential term with an arbitrary base, we can apply the substitution ๐‘Ž=๐‘’๏Œป๏—๏Œป๏—๏Œบln to derive the following general result: ๏„ธ๐‘Ž๐‘ฅ=1๐‘๐‘Ž๐‘Ž+๏Œป๏—๏Œป๏—dlnC and ๏„ธ๐‘Ž๐‘ฅ๐‘ฅ=๐‘Ž|๐‘ฅ|+.dlnC

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