Problem 33 - Calculus and its applications

Question

Evaluate. (Be sure to check by differentiating!) $$ \int \frac{d x}{x \ln x}, x>1 $$

Step-by-Step Solution

Answer

The integral is \( \ln |\ln x| + C \).

Step 1: Identify the Integral Form

We need to solve the integral \( \int \frac{dx}{x \ln x} \). This integral fits the form \( \int \frac{1}{u} \, du \) if we make an appropriate substitution.

Step 2: Choose a Suitable Substitution

Let \( u = \ln x \). Then \( du = \frac{1}{x} \, dx \). Substitute back into the integral to get \( \int \frac{dx}{x \ln x} = \int \frac{1}{u} \, du \).

Step 3: Integrate Using the Substitution

The integral \( \int \frac{1}{u} \, du \) results in \( \ln |u| + C \), where \( C \) is the constant of integration. Thus, substituting back for \( u \), we have \( \ln |\ln x| + C \).

Step 4: Differentiate to Check the Result

To verify, differentiate \( \ln |\ln x| + C \) with respect to \( x \). By the chain rule, \( \frac{d}{dx}[\ln |\ln x|] = \frac{1}{\ln x} \cdot \frac{1}{x} = \frac{1}{x \ln x} \), which matches the original integrand.

Key Concepts

Understanding u-substitution

Integration is an essential skill in calculus, and sometimes direct integration is not straightforward. This is where the u-substitution technique becomes very helpful. U-substitution is a method used to simplify integrals by making a substitution that reduces a complicated expression into a simpler form.

• First, we identify a part of the integral as a new variable, often labeled as "u." This part is usually found in the denominator or inside parentheses.
• Next, we express every part of the integral in terms of this new variable "u" and its derivative "du."
• The substituted integral can often be recognized as a basic integral form, making it easier to solve.

Consider the integral given in our exercise: \( \int \frac{dx}{x \ln x} \). Here, setting \( u = \ln x \) transforms the integration into a much simpler form \( \int \frac{1}{u} \, du \). The derivative of \( \ln x \) is \( \frac{1}{x} \), so \( du = \frac{1}{x} \, dx \). This substitution directly simplifies the task, showcasing the powerful simplicity of u-substitution.

Recognizing basic integral forms

Once a substitution reduces an integral to a simpler form, recognizing basic integral forms is the next step. Basic integral forms are standard integrals that most calculus students are familiar with, potentially allowing instant solutions without further calculation.

• A well-known basic integral form is \( \int \frac{1}{u} \, du = \ln |u| + C \), where \( C \) represents the constant of integration.
• Recognizing these forms accelerates the integration process and reduces potential errors.
• This forms the backbone of calculus and makes dealing with more complex integrals manageable by breaking them down into simpler parts.

For our integral \( \int \frac{dx}{x \ln x} \), after substituting \( u = \ln x \), the problem transforms into \( \int \frac{1}{u} \, du \), an integral form that yields \( \ln |u| + C \). Substituting back, the solution becomes \( \ln |\ln x| + C \), demonstrating that identifying the integral's shape is crucial.

Applying the chain rule for verification

The chain rule in calculus is a critical concept often used to differentiate composite functions. It is also handy for verifying the correctness of an integral solution by differentiating the result to see if it matches the original integrand.

• The chain rule states that if a function \( y = f(g(x)) \), then the derivative \( y' = f'(g(x)) \cdot g'(x) \).
• This technique ensures that our integration process was correct and can provide an intuitive understanding of how functions are related.
• In examining our result \( \ln |\ln x| + C \), applying the chain rule gives \( \frac{d}{dx}[\ln |\ln x|] = \frac{1}{\ln x} \cdot \frac{d}{dx}[\ln x] = \frac{1}{\ln x} \cdot \frac{1}{x} = \frac{1}{x \ln x} \).

Differentiating our solution recovers the original integrand, confirming that our integration and substitution were done correctly. Without this step, the solution's accuracy could remain in doubt, underscoring the importance of the chain rule in calculus.