Lesson Explainer: Integrals Resulting in Logarithmic Functions Mathematics • Higher Education

In this explainer, we will learn how to evaluate integrals of functions in the form 𝑓′(π‘₯)𝑓(π‘₯), resulting in logarithmic functions.

Integrals resulting in logarithmic functions have many real-world applications as these functions are used in mathematical models to describe population growth, cell growth or radioactive decay.

The integrals of the form 𝑓′(π‘₯)𝑓(π‘₯) are generalizations of the integral for reciprocal functions such as ο„Έ1π‘₯π‘₯=|π‘₯|+,dlnC with 𝑓(π‘₯)=π‘₯ and 𝑓′(π‘₯)=1. For a generic function 𝑓(π‘₯), we want to evaluate the integral 𝑓′(π‘₯)𝑓(π‘₯)π‘₯.d

This can be shown directly using integration by substitution. Taking the change of variable as 𝑒=𝑓(π‘₯), dd𝑒π‘₯=𝑓′(π‘₯).

We can separate the variables in this derivative by treating it like a fraction and manipulating the differentials, or more precisely, using dddd𝑒=𝑒π‘₯π‘₯.

Hence, dd𝑒=𝑓′(π‘₯)π‘₯.

Therefore, upon making the change of variable with 𝑒=𝑓(π‘₯), we have 𝑓′(π‘₯)𝑓(π‘₯)π‘₯=ο„Έο€½1𝑓(π‘₯)𝑓′(π‘₯)π‘₯=ο„Έ1𝑒𝑒=|𝑒|+=|𝑓(π‘₯)|+.dddlnClnC

We have used the integral for reciprocal functions in order to evaluate this integral. This result leads us to the following definition.

Definition: Integrals of the Form 𝑓'(π‘₯)/𝑓(π‘₯)

ο„Έπ‘Žπ‘“β€²(π‘₯)𝑓(π‘₯)π‘₯=π‘Ž|𝑓(π‘₯)|+,dlnC(1)

for some π‘Žβˆˆβ„.

We can verify this by differentiating ln|𝑓(π‘₯)| via the chain rule and demonstrating the result to be equal to the integrand of (1) or by letting 𝑓(π‘₯)=𝑒,() for some arbitrary function 𝑔(π‘₯). We can differentiate this to obtain 𝑓′(π‘₯)=𝑔′(π‘₯)𝑒=𝑔′(π‘₯)𝑓(π‘₯).()

Hence, 𝑔′(π‘₯)=𝑓′(π‘₯)𝑓(π‘₯).

But since 𝑔(π‘₯)=|𝑓(π‘₯)|ln, we find that ddlnπ‘₯(|𝑓(π‘₯)|)=𝑓′(π‘₯)𝑓(π‘₯).

The rule (1) given in the definition allows you to integrate a particular class of integrals resulting in a logarithmic function when the integrand is expressed in a special form 𝑓′(π‘₯)𝑓(π‘₯) and by appropriately identifying 𝑓(π‘₯).

Clearly this method works for integrating reciprocal functions like 1π‘₯, but what if we have higher order polynomials in the numerator or denominator? As long as we can express the derivative of the denominator as some scalar multiple of the numerator, then the method can be applied. To see this, consider the integral ο„Έ2π‘₯π‘₯+1π‘₯.d

The denominator, 𝑓(π‘₯)=π‘₯+1, can be differentiated to give 𝑓′(π‘₯)=2π‘₯, which is indeed some scalar multiple of the numerator. Therefore, from (1) we can deduce that ο„Έ2π‘₯π‘₯+1π‘₯=||π‘₯+1||+.dlnC

Now, consider the integral ο„Έ6π‘₯+16π‘₯+6π‘₯+4π‘₯+3π‘₯+7π‘₯.d

The denominator, 𝑓(π‘₯)=π‘₯+4π‘₯+3π‘₯+7, can be differentiated to give 𝑓′(π‘₯)=3π‘₯+8π‘₯+3=12ο€Ή6π‘₯+16π‘₯+6, which is, once again, a scalar multiple of the numerator. We can immediately use the definition (1) to deduce that ο„Έ6π‘₯+16π‘₯+6π‘₯+4π‘₯+3π‘₯+7π‘₯=2ο„Έ3π‘₯+8π‘₯+3π‘₯+4π‘₯+3π‘₯+7π‘₯=2||π‘₯+4π‘₯+3π‘₯+7||+.ddlnC

Some fractional expressions may appear too difficult to handle and it may not be obvious that we can use this method to integrate. However, with a bit of clever manipulation, such as multiplying the integrand by 𝑔(π‘₯)𝑔(π‘₯) for some appropriate function 𝑔(π‘₯), we may be able to put them in the particular form for the definition (1) to apply.

To see this in action, consider the integral ο„Έπ‘’βˆ’1π‘₯𝑒+1π‘₯.d

It may not be immediately obvious that the method can be applied for this integral, since the derivative of the denominator of the integrand is not a multiple of the numerator. In particular, if 𝑓(π‘₯)=π‘₯𝑒+1, then 𝑓′(π‘₯)=(π‘₯+1)𝑒, which is clearly not a scalar multiple of the numerator in the current form. However, we can manipulate the integrand by multiplying the numerator and denominator by π‘’οŠ±ο—: ο€½π‘’βˆ’1π‘₯𝑒+1𝑒𝑒=1βˆ’π‘’π‘₯+𝑒.ο—ο—οŠ±ο—οŠ±ο—οŠ±ο—οŠ±ο—

In this form, we can identify 𝑓(π‘₯)=π‘₯+𝑒,οŠ±ο— which, upon differentiating, gives 𝑓′(π‘₯)=1βˆ’π‘’.οŠ±ο—

Now, we see that 𝑓′(π‘₯) is a scalar multiple of the numerator. Applying rule (1), we find ο„Έπ‘’βˆ’1π‘₯𝑒+1π‘₯=ο„Έ1βˆ’π‘’π‘₯+𝑒π‘₯=|π‘₯+𝑒|+.ο—ο—οŠ±ο—οŠ±ο—οŠ±ο—ddlnC

Now, consider the integral ο„Έπ‘₯π‘₯(π‘₯)π‘₯.cossinlnsind

Again, it may not be immediately obvious that the method can be applied. In this form, if we take the denominator, 𝑓(π‘₯)=π‘₯(π‘₯),sinlnsin the derivative is, 𝑓′(π‘₯)=π‘₯((π‘₯)+1),coslnsin

which is not a multiple of the numerator. However, if we express cotcossinπ‘₯=π‘₯π‘₯, we can rewrite the integrand as cossinlnsincotlnsinπ‘₯π‘₯(π‘₯)=π‘₯(π‘₯).

In this form, we can identify the denominator as 𝑓(π‘₯)=(π‘₯),lnsin and the derivative as 𝑓′(π‘₯)=π‘₯,cot which is indeed a scalar multiple of the numerator. Therefore, using (1), we can deduce that ο„Έπ‘₯π‘₯(π‘₯)π‘₯=ο„Έπ‘₯(π‘₯)π‘₯=|(π‘₯)|+.cossinlnsindcotlnsindlnlnsinC

These examples illustrate the importance of having the integrand in the right form and picking the appropriate 𝑓(π‘₯) for this method to apply. Finding the right way to manipulate the expressions to get them in the right form comes from practice and experience, which is why it is a good idea to complete as many examples as possible. All of these types of integrals can also be solved using integration by substitution after choosing a suitable change of variable.

Now, let’s look at a few examples to practice and deepen our understanding of finding the integral of functions in the form 𝑓′(π‘₯)𝑓(π‘₯) using the rule (1) given in the definition.

The first example involves integrating a rational function with polynomials in the numerator and denominator.

Example 1: Finding the Integration of a Rational Function

Determine ο„Έ2π‘₯+1π‘₯+π‘₯βˆ’7π‘₯d.

Answer

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

We first identify the denominator of the integrand as 𝑓(π‘₯)=π‘₯+π‘₯βˆ’7.

We know that we can use the formula ο„Έπ‘Žπ‘“β€²(π‘₯)𝑓(π‘₯)π‘₯=π‘Ž|𝑓(π‘₯)|+,π‘Žβˆˆβ„,dlnC if the derivative of the denominator of the integrand is a scalar multiple of the numerator. In this case, we have 𝑓′(π‘₯)=2π‘₯+1.

Since this is the numerator of the integrand, we can apply the formula to find ο„Έ2π‘₯+1π‘₯+π‘₯βˆ’7π‘₯=||π‘₯+π‘₯βˆ’7||+.dlnC

Alternatively, we could have used integration by substitution with the change of variable 𝑒=π‘₯+π‘₯βˆ’7 to get the same answer.

In our next example, we will learn how to integrate a reciprocal trigonometric function by applying the rule (1).

Example 2: Finding the Integration of a Function Involving Reciprocal Trigonometric Functions

Determine ο„Έπ‘₯π‘₯cotd.

Answer

In this example, we want to find the integral of a function made up of a reciprocal trigonometric function, which can itself be represented as the quotient of two trigonometric functions. We will evaluate this integral using the formula ο„Έπ‘Žπ‘“β€²(π‘₯)𝑓(π‘₯)π‘₯=π‘Ž|𝑓(π‘₯)|+,π‘Žβˆˆβ„.dlnC

We begin by using the definition of cotπ‘₯ in terms of tanπ‘₯, which can also be expressed in terms of sinπ‘₯ and cosπ‘₯, and we can rewrite the integrand as cottancossinπ‘₯=1π‘₯=π‘₯π‘₯.

We can check that this integral is in the special form by identifying the denominator as 𝑓(π‘₯)=π‘₯,sin and checking that the derivative of this is some multiple of the numerator. Indeed, we find 𝑓′(π‘₯)=π‘₯.cos

Thus, we can use the formula to deduce that ο„Έπ‘₯π‘₯=ο„Έπ‘₯π‘₯π‘₯=|π‘₯|+.cotdcossindlnsinC

We can also find this integral using the substitution 𝑒=π‘₯sin.

In our next example, we will demonstrate how to integrate a rational function made up of the quotient of two polynomials.

Example 3: Finding the Integration of a Rational Function Using Integration by Substitution

Determine ο„Έπ‘₯+7π‘₯+21π‘₯βˆ’5π‘₯d.

Answer

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

We first identify the denominator of the integrand as 𝑓(π‘₯)=π‘₯+21π‘₯βˆ’5, and check that the derivative of this is some multiple of the numerator. Indeed, we have 𝑓′(π‘₯)=3π‘₯+21=3ο€Ήπ‘₯+7.

Since this is in a special form, ο„Έπ‘Žπ‘“β€²(π‘₯)𝑓(π‘₯)π‘₯=π‘Ž|𝑓(π‘₯)|+,π‘Žβˆˆβ„,dlnC we can evaluate the integral as ο„Έπ‘₯+7π‘₯+21π‘₯βˆ’5π‘₯=ο„Έ(3π‘₯+21)π‘₯+21π‘₯βˆ’5π‘₯=13ο„Έ3π‘₯+21π‘₯+21π‘₯βˆ’5π‘₯=13||π‘₯+21π‘₯βˆ’5||+.dddlnC

We can also evaluate this integral using the substitution 𝑒=π‘₯+21π‘₯βˆ’5.

In our next example, we will integrate a quotient of two trigonometric functions.

Example 4: Finding the Integration of a Function Involving Trigonometric Functions Using Integration by Substitution

Determine ο„Έ27π‘₯+21π‘₯7π‘₯βˆ’9π‘₯π‘₯sincossincosd.

Answer

In this example, we want to find the integral of a trigonometric function. We will evaluate this integral using the formula ο„Έπ‘Žπ‘“β€²(π‘₯)𝑓(π‘₯)π‘₯=π‘Ž|𝑓(π‘₯)|+,π‘Žβˆˆβ„.dlnC

We first identify the denominator of the integrand as 𝑓(π‘₯)=7π‘₯βˆ’9π‘₯sincos and check that the derivative of this is some scalar multiple of the numerator. Indeed, we have 𝑓′(π‘₯)=7π‘₯+9π‘₯=13(27π‘₯+21π‘₯).cossinsincos

Since this is the required form and 𝑓′(π‘₯) is a scalar multiple of the numerator of the integrand, we can deduce the integral from the formula: ο„Έ27π‘₯+21π‘₯7π‘₯βˆ’9π‘₯π‘₯=ο„Έ3(7π‘₯+9π‘₯)7π‘₯βˆ’9π‘₯π‘₯=3ο„Έ7π‘₯+9π‘₯7π‘₯βˆ’9π‘₯π‘₯=3|7π‘₯βˆ’9π‘₯|+.sincossincosdcossinsincosdcossinsincosdlnsincosC

We can also evaluate this integral using the substitution 𝑒=7π‘₯βˆ’9π‘₯sincos.

In our first few examples, we learned how to apply the special rule (1) for integrating, where the numerator is a scalar multiple of the derivative of the denominator, which results in a logarithmic function. We can also apply this rule when the integrand itself already contains logarithms. Let’s demonstrate this in our next example.

Example 5: Finding the Integration of a Function Involving Logarithmic Function Using Integration by Substitution

Determine ο„Έβˆ’3π‘₯8π‘₯π‘₯lnd.

Answer

In this example, we want to find the integral of a function involving a logarithmic function. We will evaluate this integral using the formula ο„Έπ‘Žπ‘“β€²(π‘₯)𝑓(π‘₯)π‘₯=π‘Ž|𝑓(π‘₯)|+,π‘Žβˆˆβ„.dlnC

It may not be immediately obvious how to put this integral into the special form required by this formula. For this reason, we rewrite the integrand: βˆ’3π‘₯8π‘₯=8π‘₯.lnlnοŠ±οŠ©ο—

We now identify the denominator of the integrand as 𝑓(π‘₯)=8π‘₯ln and check that the derivative of this is some multiple of the numerator. Indeed, we find 𝑓′(π‘₯)=1π‘₯=βˆ’13ο€Όβˆ’3π‘₯.

Since this is in the right form and 𝑓′(π‘₯) is a scalar multiple of the numerator of the integrand, we can deduce the integral from the formula: ο„Έβˆ’3π‘₯8π‘₯π‘₯=ο„Έ8π‘₯π‘₯=βˆ’3ο„Έ8π‘₯=βˆ’3|8π‘₯|+.lndlndlnlnlnCοŠ±οŠ©ο—οŠ§ο—

We can also evaluate this integral using the substitution 𝑒=8π‘₯ln.

In our next example, we will learn how to manipulate a reciprocal trigonometric function to get it into the form 𝑓′(π‘₯)𝑓(π‘₯) so that we can evaluate its integral.

Example 6: Integrating Reciprocal Trigonometric Functions

Determine ο„Έ27π‘₯π‘₯cscd.

Answer

In this example, we want to find the integral of a reciprocal trigonometric function. We will evaluate this integral using the formula ο„Έπ‘Žπ‘“β€²(π‘₯)𝑓(π‘₯)π‘₯=π‘Ž|𝑓(π‘₯)|+,π‘Žβˆˆβ„.dlnC

In order to get the integral into the required form for this formula, we need to multiply the denominator and numerator by cotcsc7π‘₯+7π‘₯ since the derivative of this will be a scalar multiple of the numerator. If we do this, the integrand becomes 27π‘₯ο€Ό7π‘₯+7π‘₯7π‘₯+7π‘₯=2ο€Ύ7π‘₯7π‘₯+7π‘₯7π‘₯+7π‘₯.csccotcsccotcsccsccotcsccotcsc

We now identify the denominator of the integrand as 𝑓(π‘₯)=7π‘₯+7π‘₯,cotcsc and check that the derivative of this is some multiple of the numerator. For this, we will use ddcotcscddcsccotcscπ‘₯(π‘Žπ‘₯)=βˆ’π‘Žπ‘Žπ‘₯,π‘₯(π‘Žπ‘₯)=βˆ’π‘Žπ‘Žπ‘₯π‘Žπ‘₯.

The derivative is, therefore, 𝑓′(π‘₯)=βˆ’77π‘₯7π‘₯βˆ’77π‘₯=βˆ’7ο€Ή7π‘₯7π‘₯+7π‘₯,cotcsccsccsccotcsc which is indeed a scalar factor of the numerator. Since this is in a special form, we can deduce the integral from the formula: ο„Έ27π‘₯π‘₯=2ο„Έο€Ύ7π‘₯7π‘₯+7π‘₯7π‘₯+7π‘₯π‘₯=2ο„Έο‚ο€Ήβˆ’77π‘₯7π‘₯βˆ’77π‘₯7π‘₯+7π‘₯π‘₯=βˆ’27ο„Έο€Ύβˆ’77π‘₯7π‘₯βˆ’77π‘₯7π‘₯+7π‘₯π‘₯=βˆ’27|7π‘₯+7π‘₯|+.cscdcsccotcsccotcscdcotcsccsccotcscdcotcsccsccotcscdlncotcscC

We can also evaluate this integral using the substitution 𝑒=7π‘₯+7π‘₯cotcsc.

In our last example, we will once again use trigonometric identities to help us evaluate an integral.

Example 7: Integrating Trigonometric Functions

Determine ο„Έβˆ’56π‘₯π‘₯tand.

Answer

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

By recognizing the trigonometric identity tansincosπ‘₯=π‘₯π‘₯, we can write the integrand as tansincos6π‘₯=6π‘₯6π‘₯.

We can identify the denominator of the integrand as 𝑓(π‘₯)=6π‘₯,cos which, upon differentiating, gives 𝑓′(π‘₯)=βˆ’66π‘₯.sin

Therefore, the integrand is in a special form and we can apply the formula ο„Έπ‘Žπ‘“β€²(π‘₯)𝑓(π‘₯)π‘₯=π‘Ž|𝑓(π‘₯)|+,π‘Žβˆˆβ„.dlnC

Hence, ο„Έβˆ’56π‘₯π‘₯=5ο„Έβˆ’6π‘₯π‘₯=5ο„Έβˆ’6π‘₯6π‘₯π‘₯=5ο„Έ(βˆ’66π‘₯)6π‘₯π‘₯=56ο„Έβˆ’66π‘₯6π‘₯π‘₯=56|6π‘₯|+.tandtandsincosdsincosdsincosdlncosC

We can also find this integral using the substitution 𝑒=6π‘₯cos.

Key Points

  • Integrals of the form 𝑓′(π‘₯)𝑓(π‘₯) can be evaluated using the formula ο„Έπ‘Žπ‘“β€²(π‘₯)𝑓(π‘₯)π‘₯=π‘Ž|𝑓(π‘₯)|+,π‘Žβˆˆβ„.dlnC
  • This method can be used to find the integrals of trigonometric functions such as cotπ‘₯, tanπ‘₯, cscπ‘₯, etc.
  • Sometimes it is not immediately obvious that we can use this method to evaluate the integral, but we can manipulate the integral into the right form by rewriting the integrand or multiplying it by some appropriate 𝑔(π‘₯)𝑔(π‘₯).

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