Problem 32 - Applied Calculus

Question

Use integration by parts to find each integral. \(\int(x+a)^{n} \ln (x+a) d x \quad(n \neq-1)\)

Step-by-Step Solution

Answer

\( \int (x+a)^n \ln(x+a) dx = \frac{(x+a)^{n+1}}{n+1} \ln(x+a) - \frac{(x+a)^{n+1}}{(n+1)^2} + C \).

Step 1: Identify Parts for Integration by Parts

In integration by parts, we use the formula: \( \int u \, dv = uv - \int v \, du \). Let's choose \( u = \ln(x+a) \) and \( dv = (x+a)^n \, dx \). This choice is useful because the derivative of the logarithmic function simplifies well.

Step 2: Compute du and v

First, differentiate \( u \): \( du = \frac{1}{x+a} \, dx \). Then, integrate \( dv \) to find \( v \): \( v = \frac{(x+a)^{n+1}}{n+1} \), assuming \( n eq -1 \) to avoid division by zero.

Step 3: Apply Integration by Parts Formula

Substitute \( u \), \( dv \), \( v \), and \( du \) into the integration by parts formula: \[ \int (x+a)^n \ln(x+a) \, dx = \ln(x+a) \cdot \frac{(x+a)^{n+1}}{n+1} - \int \frac{(x+a)^{n+1}}{n+1} \cdot \frac{1}{x+a} \, dx \]. Simplify the remaining integral.

Step 4: Simplify Remaining Integral

The remaining integral becomes: \( \frac{1}{n+1} \int (x+a)^{n} \, dx \), since \( \frac{(x+a)^{n+1}}{x+a} = (x+a)^n \). This simplifies to \( \frac{1}{n+1} \cdot \frac{(x+a)^{n+1}}{n+1} + C \), where \( C \) is the constant of integration.

Step 5: Combine and Simplify Results

Combine the integration by parts result and the simplified integral: \[ \int (x+a)^n \ln(x+a) \, dx = \frac{(x+a)^{n+1}}{n+1} \ln(x+a) - \frac{(x+a)^{n+1}}{(n+1)^2} + C \].

Key Concepts

Logarithmic Integration

Logarithmic functions often appear in integration problems, particularly when dealing with products involving logarithms. A key technique for integrating forms like these is the method of integration by parts. This is particularly helpful with expressions such as \( \int (x+a)^{n} \ln(x+a) \, dx \). Here, the natural logarithm \( \ln(x+a) \) is chosen strategically to represent one of the parts due to its simpler derivative. This choice facilitates the integration process by reducing complexity in subsequent steps.
To practically apply logarithmic integration using integration by parts:

• Select \( u \) as the logarithmic part \( \ln(x+a) \) because its derivative simplifies into a basic fraction, \( du = \frac{1}{x+a} \, dx \).
• The \( dv \) is chosen to be the polynomial form \( (x+a)^n \, dx \), making integration manageable.

By understanding these selections, you'll see the elegance of logarithm simplifications in integrals.

Integration Techniques

Integration by parts is one of the versatile techniques used to solve integrals involving products of functions. It's based on the product rule of differentiation and provides a systematic way to address cases like \( \int u \, dv = uv - \int v \, du \). Breaking down the integration into these components makes the task of integrating complex products much simpler.
Several key steps are essential:

• Identify appropriate parts: Choose \( u \) and \( dv \) wisely, often opting for a logarithm or polynomial that can be easily differentiated or integrated, respectively.
• Compute derivatives and integrals: Find \( du \) by differentiating \( u \) and \( v \) by integrating \( dv \), ensuring any divisions are well-defined.
• Apply and simplify: Plug these into the integration by parts formula, simplifying any remaining integrals.

A strategic choice of \( u \) and \( dv \) often determines the ease of integrating complex expressions.

Calculus Integrals

In calculus, integrals can often seem challenging, yet they are foundational in understanding areas under curves and accumulating quantities. Mastering integrals involves learning different techniques to handle the variety of forms an integral may take, such as polynomials, exponential, logarithmic, and trigonometric functions.
Key aspects of calculus integrals include:

• Basic Antiderivatives: Knowing the basic antiderivatives like \( \int x^n \, dx \), \( \int e^x \, dx \), or \( \int \ln x \, dx \).
• Advanced Methods: Techniques like substitution and integration by parts expand the possibilities for solving complex integrals.
• Concept of Constancy: Remembering to always add the constant of integration \( C \) in indefinite integrals.

By approaching each integral with the right technique and understanding, you unlock more complex mathematical models and real-world problems. The key is practicing these techniques across various types of integrals to develop a versatile problem-solving toolkit.