Solve ∫ (2x+3)/((x-5)(x+4)+14) Dx Using Partial Fractions A Step-by-Step Guide

Jul 13, 2025 by ADMIN 79 views

Integrating with Partial Fractions A Comprehensive Guide to Solving ∫ (2x+3)/((x-5)(x+4)+14) dx

Introduction

In the realm of calculus, integration stands as a fundamental operation with diverse applications across various scientific and engineering disciplines. However, not all integrals are created equal; some present a formidable challenge, demanding the application of advanced techniques. Among these techniques, partial fraction decomposition emerges as a powerful tool for simplifying integrals of rational functions – functions expressed as the ratio of two polynomials. This comprehensive guide delves into the intricacies of using partial fractions to evaluate the integral ∫ (2x+3)/((x-5)(x+4)+14) dx, providing a step-by-step approach that will equip you with the knowledge and skills to tackle similar integration problems.

Understanding Partial Fraction Decomposition

At its core, partial fraction decomposition is an algebraic technique that transforms a complex rational function into a sum of simpler fractions. This transformation is particularly useful in integration because simpler fractions are often easier to integrate than the original complex one. The fundamental principle behind this technique lies in the fact that any rational function where the degree of the numerator is less than the degree of the denominator can be expressed as a sum of fractions whose denominators are factors of the original denominator.

To effectively employ partial fraction decomposition, it's essential to grasp the different cases that arise based on the nature of the denominator's factors. These cases include:

Distinct Linear Factors: The denominator can be factored into distinct linear terms (e.g., (x-a)(x-b), where a ≠ b).
Repeated Linear Factors: The denominator contains linear factors that are repeated (e.g., (x-a)^2(x-b)).
Irreducible Quadratic Factors: The denominator includes quadratic factors that cannot be factored further into linear terms with real coefficients (e.g., x^2 + 1).
Repeated Irreducible Quadratic Factors: The denominator contains irreducible quadratic factors that are repeated (e.g., (x^2 + 1)^2).

Each case necessitates a slightly different approach in setting up the partial fraction decomposition, but the underlying principle remains the same: to express the original rational function as a sum of simpler fractions that can be readily integrated. Mastering partial fraction decomposition is not just about manipulating algebraic expressions; it's about developing a strategic mindset for problem-solving in calculus. It involves recognizing patterns, making informed decisions about which technique to apply, and executing the steps with precision. This skill is invaluable not only for solving integrals but also for tackling a wide range of mathematical and scientific challenges.

Step-by-Step Solution for ∫ (2x+3)/((x-5)(x+4)+14) dx

Now, let's embark on a detailed journey to solve the integral ∫ (2x+3)/((x-5)(x+4)+14) dx using the power of partial fraction decomposition. We'll dissect each step, providing clarity and insights to ensure a thorough understanding of the process. This step-by-step approach is not just about finding the solution; it's about developing a systematic problem-solving methodology that you can apply to a wide array of integration challenges.

Step 1: Simplify the Denominator

The first crucial step in tackling this integral is to simplify the denominator. This involves expanding the product (x-5)(x+4) and adding 14. This seemingly simple algebraic manipulation is pivotal because it transforms the denominator into a more recognizable form, setting the stage for subsequent factorization and partial fraction decomposition.

Expanding (x-5)(x+4), we get:

(x-5)(x+4) = x^2 + 4x - 5x - 20 = x^2 - x - 20

Now, adding 14 to this result, we obtain:

x^2 - x - 20 + 14 = x^2 - x - 6

Thus, the denominator simplifies to x^2 - x - 6. This simplified form is significantly easier to work with and allows us to proceed with factoring, a critical step in the partial fraction decomposition process. The ability to simplify complex expressions is a cornerstone of mathematical proficiency, and this step exemplifies the importance of mastering algebraic manipulations in calculus.

Step 2: Factor the Denominator

The next pivotal step is to factor the simplified denominator, x^2 - x - 6. Factoring is the process of expressing a polynomial as a product of simpler polynomials. In the context of partial fraction decomposition, factoring the denominator is crucial because it reveals the linear or quadratic factors that will form the basis of our partial fractions.

To factor x^2 - x - 6, we seek two numbers that multiply to -6 and add to -1 (the coefficient of the x term). These numbers are -3 and 2. Therefore, we can factor the denominator as follows:

x^2 - x - 6 = (x - 3)(x + 2)

This factorization is a key breakthrough. It reveals that the denominator can be expressed as a product of two distinct linear factors, (x - 3) and (x + 2). This realization dictates the form of the partial fraction decomposition, as we will see in the next step. Factoring is a fundamental skill in algebra and calculus, and its mastery is essential for success in a wide range of mathematical problems. It requires a keen eye for patterns and the ability to apply various factoring techniques, such as recognizing difference of squares, perfect square trinomials, and factoring by grouping.

Step 3: Set Up the Partial Fraction Decomposition

With the denominator factored as (x - 3)(x + 2), we can now set up the partial fraction decomposition. This is where we express the original rational function as a sum of simpler fractions, each with one of the factors of the denominator as its denominator. The key principle here is that for each distinct linear factor in the denominator, we introduce a term with a constant numerator. This step is crucial because it transforms the complex rational function into a form that is amenable to integration.

Since we have two distinct linear factors, (x - 3) and (x + 2), we set up the decomposition as follows:

(2x + 3) / ((x - 3)(x + 2)) = A / (x - 3) + B / (x + 2)

Here, A and B are constants that we need to determine. This equation expresses the original rational function as a sum of two simpler fractions, each with a linear denominator. The numerators, A and B, are constants because the degree of the numerator in each partial fraction must be less than the degree of the denominator. Setting up the partial fraction decomposition correctly is paramount to solving the integral. It requires a clear understanding of the relationship between the factors of the denominator and the form of the partial fractions. This step is not just about writing down an equation; it's about translating the algebraic structure of the rational function into a form that facilitates integration.

Step 4: Solve for the Unknown Constants

The next critical step is to solve for the unknown constants, A and B, in the partial fraction decomposition. This is an algebraic process that involves manipulating the equation we set up in the previous step to isolate and determine the values of A and B. There are several methods for solving for these constants, but the most common and efficient approach is to multiply both sides of the equation by the common denominator and then strategically substitute values of x that eliminate one constant at a time. This step is pivotal because it transforms the equation into a system of linear equations that we can solve using algebraic techniques.

Multiplying both sides of the equation (2x + 3) / ((x - 3)(x + 2)) = A / (x - 3) + B / (x + 2) by (x - 3)(x + 2), we get:

2x + 3 = A(x + 2) + B(x - 3)

Now, we can substitute values of x to eliminate one constant at a time.

Let x = 3: 2(3) + 3 = A(3 + 2) + B(3 - 3) 9 = 5A A = 9/5
Let x = -2: 2(-2) + 3 = A(-2 + 2) + B(-2 - 3) -1 = -5B B = 1/5

Thus, we have found that A = 9/5 and B = 1/5. This determination of the constants is a crucial milestone in the partial fraction decomposition process. It allows us to rewrite the original rational function as a sum of simpler fractions with known coefficients, paving the way for straightforward integration. Solving for the constants requires a solid grasp of algebraic techniques, including equation manipulation and substitution. It also demands careful attention to detail to avoid errors in arithmetic, as even a small mistake can lead to an incorrect result.

Step 5: Integrate the Partial Fractions

With the constants A and B determined, we can now proceed to the integration step. This is where the true power of partial fraction decomposition becomes evident, as we transform the integral of a complex rational function into the sum of integrals of simpler fractions. The key here is that the integrals of these simpler fractions are typically straightforward and can be evaluated using standard integration techniques. This step is the culmination of our efforts, where we finally arrive at the solution to the original integral.

Substituting the values of A and B into our partial fraction decomposition, we have:

(2x + 3) / ((x - 3)(x + 2)) = (9/5) / (x - 3) + (1/5) / (x + 2)

Now, we can integrate both sides with respect to x:

∫ (2x + 3) / ((x - 3)(x + 2)) dx = ∫ [(9/5) / (x - 3) + (1/5) / (x + 2)] dx

We can split the integral into two separate integrals:

∫ [(9/5) / (x - 3) + (1/5) / (x + 2)] dx = (9/5) ∫ 1 / (x - 3) dx + (1/5) ∫ 1 / (x + 2) dx

These integrals are standard logarithmic integrals:

(9/5) ∫ 1 / (x - 3) dx = (9/5) ln|x - 3| + C1

(1/5) ∫ 1 / (x + 2) dx = (1/5) ln|x + 2| + C2

Combining these results, we get:

∫ (2x + 3) / ((x - 3)(x + 2)) dx = (9/5) ln|x - 3| + (1/5) ln|x + 2| + C

where C = C1 + C2 is the constant of integration. This final expression represents the solution to the original integral. The process of integrating the partial fractions is often the most straightforward part of the entire process, but it relies heavily on the accuracy of the preceding steps. It requires a firm understanding of basic integration rules, particularly the integration of functions of the form 1/x, which leads to logarithmic expressions. The constant of integration, C, is a crucial reminder that the indefinite integral represents a family of functions that differ by a constant.

Conclusion

In conclusion, the evaluation of ∫ (2x+3)/((x-5)(x+4)+14) dx exemplifies the power and elegance of partial fraction decomposition in tackling complex integrals. By systematically simplifying the integrand, factoring the denominator, setting up the partial fraction decomposition, solving for the unknown constants, and integrating the resulting simpler fractions, we successfully navigated this integration challenge. This step-by-step approach not only provides the solution to this specific problem but also equips you with a robust methodology for tackling a wide range of integration problems involving rational functions.

Mastering partial fraction decomposition is an invaluable asset for anyone pursuing studies in mathematics, physics, engineering, or any field that relies heavily on calculus. It's a testament to the interconnectedness of mathematical concepts, bridging algebra and calculus in a seamless manner. By developing a deep understanding of this technique, you'll not only enhance your problem-solving skills but also gain a greater appreciation for the beauty and power of mathematical reasoning. Remember, the journey of learning calculus is not just about memorizing formulas and procedures; it's about cultivating a strategic mindset, developing a keen eye for patterns, and fostering the ability to approach complex problems with confidence and clarity. So, embrace the challenge, practice diligently, and let the power of partial fraction decomposition unlock new horizons in your mathematical endeavors.