Problem 21 - Calculus Early Transcendentals: Pearson New International Edition

Question

Use the method of partial fraction decomposition to perform the required integration. \(\int \frac{x+1}{(x-3)^{2}} d x\)

Step-by-Step Solution

Answer

\( \ln|x-3| - \frac{4}{x-3} + C \)

Step 1: Understand the Integral

The integral given is \( \int \frac{x+1}{(x-3)^{2}} \, dx \). Our task is to use partial fraction decomposition to simplify this expression before integration.

Step 2: Set Up Partial Fraction Decomposition

Since the denominator \( (x-3)^2 \) is a repeated linear factor, we propose the partial fraction decomposition as \( \frac{x+1}{(x-3)^2} = \frac{A}{x-3} + \frac{B}{(x-3)^2} \).

Step 3: Clear Denominators

Multiply through by \( (x-3)^2 \) to eliminate the denominator: \( x+1 = A(x-3) + B \).

Step 4: Expand and Collect Terms

Expand the right side: \( A(x-3) + B = Ax - 3A + B \). Collect like terms: \( x + 1 = Ax - 3A + B \).

Step 5: Solve for Coefficients

By comparing coefficients of terms, set up the equations: \( A = 1 \) and \( -3A + B = 1 \). Substitute \( A = 1 \) into the second equation to find \( B = 4 \).

Step 6: Rewrite the Integral

Substitute the values of \( A \) and \( B \) into the partial fraction form to get \( \int \left( \frac{1}{x-3} + \frac{4}{(x-3)^2} \right) \ dx \).

Step 7: Integrate Each Term

Integrate each term separately: \( \int \frac{1}{x-3} \, dx = \ln|x-3| + C_1 \) and \( \int \frac{4}{(x-3)^2} \, dx = -\frac{4}{x-3} + C_2 \).

Step 8: Combine Results

Combine the results to get \( \ln|x-3| - \frac{4}{x-3} + C \), where \( C \) is the constant of integration.

Key Concepts

Integration Techniques

When solving integrals, sometimes the function can be quite complex, necessitating a variety of methods to simplify or directly integrate the expression. One such technique is **partial fraction decomposition**. This method is particularly useful when dealing with rational functions, which are fractions where the numerator and/or the denominator is a polynomial.

• **Rational Functions:** Look like \( \frac{P(x)}{Q(x)} \), where \( P(x) \) and \( Q(x) \) are polynomials.
• **Partial Fraction Decomposition:** The aim is to express the original complicated fraction as a sum of simpler fractions that are easier to integrate.

This process typically involves setting up equations based on the different factors of the denominator, finding constants (like \( A \) and \( B \) in our example), and then integrating each fraction individually. It's a powerful method because once the fraction is decomposed, the integration usually becomes straightforward by using basic integration rules for simpler functions.

Calculus

Calculus is a branch of mathematics that is all about understanding how things change or accumulate. This problem involves integration, which is essentially a method for adding up infinitely many small quantities.
In depth, the integral \( \int \frac{x+1}{(x-3)^2} \, dx \) is tackled by:

• **Finding Antiderivatives:** This means determining a function whose derivative gives back the original function inside the integral.
• **Evaluating Bounds:** Though not required for indefinite integrals, this is typically the next step for definite integrals, which result in a specific number instead of a formula.

In our example, through partial fraction decomposition, it becomes much easier to find antiderivatives because each part of the decomposed fractions corresponds to basic functions with known integrals.
Think of calculus as a toolkit. It allows mathematicians and scientists to model and solve real-world problems involving rate of change and accumulation of quantities, such as velocity and area beneath curves.

Repeated Linear Factors

When dealing with rational functions, understanding the behavior and handling of repeated linear factors is crucial, especially during partial fraction decomposition.

In the context of the given integral, the repeated factor was \((x-3)^2\). Here's how handling repeated factors differs from simple distinct ones:

• **Setup of Decomposition:** For a repeated linear factor like \((x-3)^2\), the decomposition should include not just \( \frac{A}{x-3} \) but also \( \frac{B}{(x-3)^2} \) to address each power of the factor.
• **Solving for Constants:** Equating the original function to its decomposed form helps in solving for unknown constants by generating a system of equations. In the outlined solution, coefficients were determined, which enabled rewriting of the integral in simpler terms.

Repeated linear factors require solving for more coefficients compared to distinct factors, adding a layer of complexity. Successfully decomposing these enables easier integration of each term thanks to simpler forms that can be integrated using basic rules of calculus.