Problem 31 - Calculus Early Transcendentals: Pearson New International Edition

Question

Use the method of partial fraction decomposition to perform the required integration. \(\int \frac{1}{(x-1)^{2}(x+4)^{2}} d x\)

Step-by-Step Solution

Answer

Decompose the fraction, solve for constants, then integrate each term.

Step 1: Identify the Form of Partial Fractions

First, we recognize that our integrand \( \frac{1}{(x-1)^{2}(x+4)^{2}} \) can be decomposed using partial fractions. Since the factors are \((x-1)^2\) and \((x+4)^2\), we'll set up the partial fraction decomposition as follows:\[\frac{1}{(x-1)^{2}(x+4)^{2}} = \frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{C}{x+4} + \frac{D}{(x+4)^2}\]

Step 2: Clear the Denominator

Multiply both sides of the equation by \((x-1)^2(x+4)^2\) to eliminate the denominator:\[1 = A(x-1)(x+4)^2 + B(x+4)^2 + C(x-1)^2(x+4) + D(x-1)^2\]

Step 3: Expand and Combine Terms

Expand the right side and collect terms according to powers of \(x\). Set the coefficients equal to 0 to form a system of equations, allowing us to solve for \(A\), \(B\), \(C\), and \(D\).

Step 4: Solve for Constants

Solving the system of equations obtained from comparing coefficients, we find the values of \(A\), \(B\), \(C\), and \(D\).

Step 5: Integrate Each Term Separately

Integrate each term of the partial fraction decomposition separately:\[\int \frac{A}{x-1} \, dx + \int \frac{B}{(x-1)^2} \, dx + \int \frac{C}{x+4} \, dx + \int \frac{D}{(x+4)^2} \, dx\]

Step 6: Compute the Integrals

Find the antiderivatives:- \( \int \frac{A}{x-1} \, dx = A \ln|x-1| + C_1 \)- \( \int \frac{B}{(x-1)^2} \, dx = -\frac{B}{x-1} + C_2 \)- \( \int \frac{C}{x+4} \, dx = C \ln|x+4| + C_3 \)- \( \int \frac{D}{(x+4)^2} \, dx = -\frac{D}{x+4} + C_4 \)

Step 7: Combine Integrals for the Final Solution

Combine the results of each integral to write the final solution. Assume the constant of integration \(C = C_1 + C_2 + C_3 + C_4\), now include this constant in your final answer.

Key Concepts

Integration Techniques

When solving calculus problems, especially those involving complicated functions like rational functions, we often rely on specific integration techniques. One of these techniques is partial fraction decomposition. This method simplifies the integrand, transforming a complex fraction into a sum of simpler fractions. This makes it easier to integrate each term separately.

Partial fraction decomposition is powerful because it breaks down a fraction into smaller, more manageable pieces. Once decomposed, we integrate term by term, utilizing basic integration rules that we are familiar with, such as integrating polynomials and logarithmic functions. This strategy speeds up calculus problem solving and allows us to understand and calculate the integral more effectively.

Rational Functions

Rational functions are expressions of the form \( \frac{P(x)}{Q(x)} \), where both \(P(x)\) and \(Q(x)\) are polynomials. In many calculus problems, particularly in integration, we encounter these functions. They are common because they often represent real-world scenarios in physics, engineering, and other scientific fields.

The goal with rational functions in integration is often to simplify the process using partial fraction decomposition. By expressing a complicated rational function as a collection of simpler fractions, we make the integration process much more manageable. This involves identifying the distinct linear and quadratic factors within the function's denominator and placing corresponding constants in their numerators. Once simplified, each fraction can be integrated individually.

Calculus Problem Solving

Calculus problem solving is about applying the right techniques at the right time. It involves a variety of strategies, from differentiation to integration, and partial fraction decomposition is one such tool. Effective problem solving requires recognizing when a particular method can simplify the ‘problem at hand’.

In the given exercise, partial fraction decomposition was used to handle the integration of a rational function. Initially, it was necessary to set up an equation that represented the integrand in terms of its partial fractions. This step was essential for simplifying the complex rational function. Next, by solving for the constants involved, each part of the decomposition could be integrated separately, using standard techniques. Finally, all the integrated parts were combined to provide a complete solution, often finishing with the inclusion of a constant of integration.

Understanding and mastering these techniques can make tackling calculus problems more straightforward, especially when dealing with integration of rational functions.