Question

$$ \text { use integration by parts to evaluate each integral. } $$ $$ \int \frac{\ln x}{x^{2}} d x $$

Short answer

The integral is \(-\frac{\ln x}{x} + \frac{1}{x} + C\).

Step-by-step solution

Identify parts for integration by parts

According to the integration by parts formula, which is \( \int u \, dv = uv - \int v \, du \), we need to identify \( u \) and \( dv \). Let \( u = \ln x \) which implies \( du = \frac{1}{x} \, dx \). Then, let \( dv = \frac{1}{x^2} \, dx \), which implies that \( v = \int \frac{1}{x^2} \, dx = -\frac{1}{x} \).

Apply the integration by parts formula

Substitute \( u \), \( v \), \( du \), and \( dv \) into the integration by parts formula: \[ \int \frac{\ln x}{x^2} \ dx = \left( \ln x \right) \left( -\frac{1}{x} \right) - \int \left( -\frac{1}{x} \right) \left( \frac{1}{x} \right) \ dx. \] Simplify this to: \[ -\frac{\ln x}{x} + \int \frac{1}{x^2} \ dx. \]

Integrate the simple integral

Now, focus on the remaining integral: \[ \int \frac{1}{x^2} \ dx = \int x^{-2} dx. \] The integral of \( x^{-2} \) is \(-x^{-1} = -\frac{1}{x} \), so we find: \[ \int \frac{1}{x^2} \, dx = -\frac{1}{x}. \]

Combine the results

Combine the results of integration by parts and the evaluated integral: \[ -\frac{\ln x}{x} - \left( -\frac{1}{x} \right) = -\frac{\ln x}{x} + \frac{1}{x}. \] Hence, the solution to the integral is: \[ -\frac{\ln x}{x} + \frac{1}{x} + C, \] where \( C \) is the integration constant.

Key concepts

Natural Logarithm

The natural logarithm, commonly denoted as \( \ln x \), is an essential function in calculus. It represents the logarithm to the base \( e \), where \( e \) is an irrational number approximately equal to 2.71828. The function \( \ln x \) is useful for modeling exponential growth and decay problems.

• It is defined only for positive numbers, which means \( x > 0 \).
• The derivative of \( \ln x \) is \( \frac{1}{x} \), which signifies its rate of change.
• The integral of \( \ln x \) is more complex and often requires integration techniques like integration by parts.

Understanding the properties of the natural logarithm is crucial because it frequently appears in calculus, especially when dealing with exponential growth, compound interest, and continuous compound interest scenarios.

Definite Integral

A definite integral, not to be confused with an indefinite integral, is a fundamental concept in calculus. It represents the area under a curve described by a function between two points \( a \) and \( b \). This is written as \( \int_{a}^{b} f(x) \, dx \).

• The result of a definite integral is a number, representing a total quantity.
• It provides a way to calculate things like areas, volumes, and other quantities that accumulate over an interval.
• Boundary points \( a \) and \( b \) are crucial as they determine the region over which we are integrating.

In problems involving definite integrals, it's important to correctly interpret the limits and understand how they limit the scope of the integration process to a finite region or time interval.

Integration Techniques

Integration by parts is one of the powerful techniques used to integrate products of functions. The fundamental formula for integration by parts is given by:\[\int u \, dv = uv - \int v \, du\]where \( u \) and \( dv \) are parts of the integrand that you choose. Here's how it works:

• Choose \( u \) to be the part of the integrand whose derivative \( du \) is simpler.
• Choose \( dv \) such that \( v \) can be easily integrated.
• Apply the integration by parts formula.
• Sometimes the resulting integral \( \int v \, du \) may need further integration techniques to solve.

In the given problem, \( u = \ln x \) and \( dv = \frac{1}{x^2} \, dx \) were chosen strategically to simplify the integration process. This technique is particularly useful when tackling integrals involving logarithmic or inverse trigonometric functions.

Calculus Problem Solving

Solving calculus problems often requires a structured approach, utilizing various techniques as needed. Here are general steps to help with calculus problem solving:

• Understand the problem: Clearly define what is being asked. Identify key functions and limits.
• Select the appropriate technique: Choose methods like integration by parts, substitution, or partial fractions.
• Apply the technique: Re-work the function based on your selected method. Double-check transformations and limits.
• Combine results: Assemble your results from different techniques, if multiple were used.
• Verify your answer: Use derivatives or numerical checks to ensure that your solution is correct.

This approach ensures a systematic search for a solution, reducing chances of errors and enhancing understanding of complex calculus problems.