# How to Antidifferentiate Logarithmic Functions

Instructor: Christopher Haines
In this lesson, we demonstrate how to evaluate four types of integrals involving a logarithmic function. The method we use for each is Integration by Parts, and one of the key tricks is realizing a constant function is the derivative of a linear function. The natural logarithm works well with this since its derivative is 1/x.

### Opening Example

In this lesson, we will be working with integrals having a logarithmic function appearing somewhere in the integrand. For example, consider the integral:

We can evaluate this integral using Integration by Parts. We pause here to remind the reader of this method.

#### Integration by Parts Formula

Suppose u and v are differentiable functions. Then,

So to evaluate this integral, we can let v(x) = ln(x) and u(x) = x. This way, the v'u integrand becomes the constant 1, since v'(x) = 1/x. The result is then

Integration by Parts can also be used to treat integrals of the following four forms.

In the sections to follow, we will learn how to evaluate each of these types of integrals.

### Integrals of the Form (1)

Form (1) is

and the power on x is any real number. For reasons which will become apparent, we will handle the case of the power on the x equaling -1 separate from its contrary case.

In this case, the integral becomes:

We can evaluate this integral using Integration by Parts or by a u-substitution. For the sake of demonstrating that it does not matter which method we choose, we will show both methods.

To set up Integration by Parts, let u'(x) = 1/x, and v(x) = ln(x). Here we let the v function be the logarithmic function because in the latter integral, it will turn in to a 1/x upon taking its derivative. When doing this, we arrive at:

Notice that the integral we are trying to find appears on both sides of the equation. Adding that integral to both sides of the equation and then solving for the integral, we obtain:

To use u-substitution, let u(x) = ln(x). Then,

We now advance to the second case.

Once again, since the derivative of ln(x) turns in to a 1/x when differentiated, we make that function the one to be differentiated. The other function we can integrate by using the Power Rule for integration.

Carrying out Integration by Parts in this manner, we obtain:

### Integrals of the Form (2)

Form (2) is:

where n is a non-negative integer.

Recall from our earlier example,

So now consider the case of n = 2. The integral is then:

Since the integrand is a product of two functions, Integration by Parts applies, and we can let:

Using the now known anti-derivative of ln(x), Integration by Parts shows that:

By again using the anti-derivative formula for ln(x) to evaluate the second integral, we arrive at:

Then simplifying the right side of the last equation, we obtain:

Next by using the formula we just derived for the case of n = 2, we can evaluate the integral:

The integrand is again a product of two functions. Let:

Then Integration by Parts results in:

The last integral is a linear combination of integrals we have already handled. By using the formulas for the integral of the first and second power of the natural logarithm and then substituting them in to that last equation, we can arrive at the integral for the third power of the natural logarithm. After performing the remaining integration and then simplifying algebraically, we obtain the anti-derivative for the third power of the natural logarithm.

Generally when we have form (2), we can break the natural logarithm in to a product of n - 1 logs and a product of one log. Then let v(x) = ln(x) and the u'(x) be the product of the other n - 1 logs and proceed to evaluate according to the Integration by Parts formula. We invite the reader to derive a general expression valid for all non-negative integers n in one of the exercises.

### Integrals of the Form (3)

A third form we show is:

Note that we have really already taken care of the case of a = 0. For when a = 0, the integral reduces to:

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