Problem 55 - University Calculus: Early Transcendentals

Question

Evaluate the integrals. $$\int \frac{d x}{x \ln x}$$

Step-by-Step Solution

Answer

The integral evaluates to \( \ln |\ln x| + C \).

Step 1: Integrate by Substitution

Let's take a substitution to simplify the integral. We choose the substitution: \( u = \ln x \). Then, the derivative of \( u \) with respect to \( x \) is \( \frac{du}{dx} = \frac{1}{x} \), which implies \( du = \frac{1}{x} dx \). This leads to the change in variables: \( dx = x \, du \). Substitute these into the integral, which gives: \[ \int \frac{d x}{x \ln x} = \int \frac{1}{u} \, du \].

Step 2: Integrate with respect to u

The integral \( \int \frac{1}{u} \, du \) is a standard integral with the solution being the natural logarithm function. So, we have \[ \int \frac{1}{u} \, du = \ln |u| + C \] where \( C \) is the integration constant.

Step 3: Substitute back the original variable

Recall the substitution \( u = \ln x \). Plug this back into the result from Step 2 to revert to the original variable: \[ \ln |u| + C = \ln |\ln x| + C \].

Step 4: Finalize the Integral

Thus, the evaluated integral in terms of \( x \) is \[ \int \frac{d x}{x \ln x} = \ln |\ln x| + C \], where \( C \) is the constant of integration.

Key Concepts

Natural Logarithm

The natural logarithm, often denoted as \( \ln(x) \), is a logarithm with base \( e \), where \( e \approx 2.71828 \). It is the inverse function of the exponential function \( e^x \). This particular logarithm is prominent in calculus due to its natural properties, making it easier to differentiate and integrate.

• When you see \( \ln(x) \), think about how it can simplify expression, especially in integration and differentiation scenarios.
• The derivative of \( \ln(x) \) is \( \frac{1}{x} \), which gets useful when you perform substitution during integration, as in the given exercise.

Understanding the behavior of the natural logarithm in calculus is crucial, as it frequently appears in solving integrals and derivatives due to its smooth integration properties with other elementary functions.

Integral Calculus

Integral calculus is a fundamental branch of calculus focused on the concept of integration, which is essentially the reverse operation of differentiation. Integrals are used to determine areas under curves, accumulate quantities, and solve complex differential equations.

• In the given exercise, we're tasked with evaluating the integral \( \int \frac{d x}{x \ln x} \), which requires transforming the expression to make it integrable.
• Integral calculus provides tools like substitution, which allows us to convert the integral into a more manageable form. Substitution often picks a part of the integrand that simplifies the expression once substituted.
• The standard form of integrals, such as \( \int \frac{1}{u} \, du = \ln |u| + C \), shows how fundamental functions behave under integration.

Comprehending the basics of integration helps in navigating through complex problems, allowing you to identify strategies for handling non-standard integrals.

Change of Variables

The change of variables technique, also known as substitution, is a powerful tool in integral calculus. This method simplifies an integral by transforming the complicated integrand into a standard form. In essence, you adjust the integration variable to reveal the function's innate simplicity.

• In the exercise, the substitution \( u = \ln x \) was chosen strategically to simplify the integral. This turns our original problem into a standard form \( \int \frac{1}{u} \, du \).
• The differential \( dx \) in terms of the new variable becomes \( x \, du \), allowing you to seamlessly switch between variables without losing the essence of the original function.
• Once you find the integral in terms of \( u \), you must substitute back the original variable to complete the solution, ensuring it reflects the initial problem statement.

By mastering the change of variables, you gain a strategic skill to tackle integrals that seem daunting at first glance, making variable replacement a valuable ally in mathematics.