Problem 33 - Applied Calculus

Question

Use integration by parts to find each integral. \(\int \ln x d x\)

Step-by-Step Solution

Answer

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

Step 1: Identify the Parts for Integration by Parts Formula

Integration by parts is based on the formula \( \int u \, dv = uv - \int v \, du \). For the integral \( \int \ln x \, dx \), we can choose \( u = \ln x \) and \( dv = dx \). This choice is strategic because the derivative of \( \ln x \) is simpler.

Step 2: Differentiate and Integrate

Differentiate \( u \) to find \( du \), and integrate \( dv \) to find \( v \):\(du = \frac{1}{x} \, dx\)\(v = \int dv = \int dx = x\).

Step 3: Apply Integration by Parts Formula

Substitute \( u \), \( du \), \( v \), and \( dv \) into the integration by parts formula:\(\int \ln x \, dx = uv - \int v \, du \)Using the previously found parts:\(= x \ln x - \int x \left( \frac{1}{x} \right) dx \).

Step 4: Simplify the Expression

Simplify the remaining integral:\(= x \ln x - \int 1 \, dx \).Therefore, \( \int 1 \, dx \) simplifies to \( x \).

Step 5: Final Integration Result

Combine and simplify the terms:\(x \ln x - x + C \), where \( C \) is the constant of integration.

Key Concepts

Logarithmic Integration

Logarithmic integration focuses on integrating functions that involve natural logarithms, such as \( \ln x \). These types of integrals often require specific techniques to simplify the integration process. Here, the focus is on finding \( \int \ln x \, dx \), a common challenge in calculus.The direct integration of \( \ln x \) is not straightforward. Using integration by parts becomes necessary to find a solution. By selecting \( u = \ln x \) and \( dv = dx \), we utilize the method effectively, as taking the derivative of the logarithmic function is usually simpler and easily manageable. This choice reflects the strategy behind logarithmic integration.Once the parts are assigned, the problem transforms and allows us to address the integral of \( \ln x \). Utilizing the properties of logarithms in conjunction with integration by parts simplifies this process. When broken down into manageable steps, logarithmic integration highlights both creativity and strategy in calculus problem-solving.

Integration Techniques

Integration techniques include a variety of methods used to solve integrals, whether definite or indefinite. One such technique, integration by parts, relies on the formula:

• \( \int u \, dv = uv - \int v \, du \)

This method involves selecting two components, \( u \) and \( dv \), from the original integral. For \( \int \ln x \, dx \), the choice of \( u = \ln x \) and \( dv = dx \) is essential.After identifying \( u \) and \( dv \), differentiating and integrating these respects provide the expressions for \( du \) and \( v \). It requires finding \( du = \frac{1}{x} \, dx \) and \( v = x \). These expressions are then used in the integration by parts formula to simplify and solve the integral.Integration techniques like this one are crucial in calculus, offering varied methods to tackle complex integrals. Picking the right technique often defines the success in finding solutions, demonstrating the importance of understanding the strengths and limitations of each method.

Calculus Problem Solving

Calculus problem solving, especially in integration, demands a strategic approach. While solving \( \int \ln x \, dx \), choosing the correct methods, like integration by parts, showcases the application of learned techniques. This structured method not only solves the integral but also aids in expanding the understanding of calculus.The problem incorporates simplifying expressions and combining them to achieve the final solution. For example, after applying the integration by parts formula, the solution requires simplifying:

• \( x \ln x - \int 1 \, dx \)

This simplification leads to a straightforward result: \( x \ln x - x + C \), showing the integral's solution.In calculus, problem solving involves using known methods innovatively while understanding the underlying principles. Through practice, recognizing when to use certain techniques, such as in logarithmic integration, becomes intuitive. Each problem presents an opportunity to develop a deeper comprehension of mathematical concepts, enhancing problem-solving skills.