Lesson Explainer: Integrals Resulting in Logarithmic Functions | Nagwa Lesson Explainer: Integrals Resulting in Logarithmic Functions | Nagwa

# Lesson Explainer: Integrals Resulting in Logarithmic Functions Mathematics

In this explainer, we will learn how to evaluate integrals of functions in the form , 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 are generalizations of the integral for reciprocal functions such as with and . For a generic function , we want to evaluate the integral

This can be shown directly using integration by substitution. Taking the change of variable as ,

We can separate the variables in this derivative by treating it like a fraction and manipulating the differentials, or more precisely, using

Hence,

Therefore, upon making the change of variable with , we have

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 π'(π₯)/π(π₯)

 οΈππβ²(π₯)π(π₯)π₯=π|π(π₯)|+,dlnC (1)

for some .

We can verify this by differentiating via the chain rule and demonstrating the result to be equal to the integrand of (1) or by letting for some arbitrary function . We can differentiate this to obtain

Hence,

But since , we find that

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 and by appropriately identifying .

Clearly this method works for integrating reciprocal functions like , 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

The denominator, can be differentiated to give which is indeed some scalar multiple of the numerator. Therefore, from (1) we can deduce that

Now, consider the integral

The denominator, can be differentiated to give which is, once again, a scalar multiple of the numerator. We can immediately use the definition (1) to deduce that

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 for some appropriate function , 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

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 then 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 :

In this form, we can identify which, upon differentiating, gives

Now, we see that is a scalar multiple of the numerator. Applying rule (1), we find

Now, consider the integral

Again, it may not be immediately obvious that the method can be applied. In this form, if we take the denominator, the derivative is,

which is not a multiple of the numerator. However, if we express , we can rewrite the integrand as

In this form, we can identify the denominator as and the derivative as which is indeed a scalar multiple of the numerator. Therefore, using (1), we can deduce that

These examples illustrate the importance of having the integrand in the right form and picking the appropriate 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, letβs look at a few examples to practice and deepen our understanding of finding the integral of functions in the form 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 .

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

We first identify the denominator of the integrand as

We know that we can use the formula if the derivative of the denominator of the integrand is a scalar multiple of the numerator. In this case, we have

Since this is the numerator of the integrand, we can apply the formula to find

Alternatively, we could have used integration by substitution with the change of variable 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 .

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

We begin by using the definition of in terms of , which can also be expressed in terms of and , and we can rewrite the integrand as

We can check that this integral is in the special form by identifying the denominator as and checking that the derivative of this is some multiple of the numerator. Indeed, we find

Thus, we can use the formula to deduce that

We can also find this integral using the substitution .

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 .

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

We first identify the denominator of the integrand as and check that the derivative of this is some multiple of the numerator. Indeed, we have

Since this is in a special form, we can evaluate the integral as

We can also evaluate this integral using the substitution .

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 .

In this example, we want to find the integral of a trigonometric function. We will evaluate this integral using the formula

We first identify the denominator of the integrand as and check that the derivative of this is some scalar multiple of the numerator. Indeed, we have

Since this is the required form and is a scalar multiple of the numerator of the integrand, we can deduce the integral from the formula:

We can also evaluate this integral using the substitution .

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. Letβs demonstrate this in our next example.

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

Determine .

In this example, we want to find the integral of a function involving a logarithmic function. We will evaluate this integral using the formula

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:

We now identify the denominator of the integrand as and check that the derivative of this is some multiple of the numerator. Indeed, we find

Since this is in the right form and is a scalar multiple of the numerator of the integrand, we can deduce the integral from the formula:

We can also evaluate this integral using the substitution .

In our next example, we will learn how to manipulate a reciprocal trigonometric function to get it into the form so that we can evaluate its integral.

### Example 6: Integrating Reciprocal Trigonometric Functions

Determine .

In this example, we want to find the integral of a reciprocal trigonometric function. We will evaluate this integral using the formula

In order to get the integral into the required form for this formula, we need to multiply the denominator and numerator by since the derivative of this will be a scalar multiple of the numerator. If we do this, the integrand becomes

We now identify the denominator of the integrand as and check that the derivative of this is some multiple of the numerator. For this, we will use

The derivative is, therefore, which is indeed a scalar factor of the numerator. Since this is in a special form, we can deduce the integral from the formula:

We can also evaluate this integral using the substitution .

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

### Example 7: Integrating Trigonometric Functions

Determine .

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

By recognizing the trigonometric identity , we can write the integrand as

We can identify the denominator of the integrand as which, upon differentiating, gives

Therefore, the integrand is in a special form and we can apply the formula

Hence,

We can also find this integral using the substitution .

### Key Points

• Integrals of the form can be evaluated using the formula
• This method can be used to find the integrals of trigonometric functions such as , , , 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 .