Partial Fraction Decomposition Of 1/((x+5)^2(x-1)) Dx

Jul 12, 2025 by ADMIN 54 views

In calculus, integration is a fundamental operation, and many techniques exist to solve various types of integrals. One such technique is the method of partial fraction decomposition, which is particularly useful for integrating rational functions. A rational function is a function that can be expressed as the quotient of two polynomials. When the denominator of a rational function can be factored, partial fraction decomposition allows us to break down the complex fraction into simpler fractions that are easier to integrate. This article will delve into the process of using partial fraction decomposition to evaluate the integral of the rational function ${ f(x) = \int \frac{1}{(x+5)^2(x-1)} dx }$ . We will explore the steps involved in decomposing the rational function and identify the factors of each partial fraction. By understanding this method, you can tackle a wide range of integrals involving rational functions more effectively.

The method of partial fraction decomposition is a powerful technique used in calculus to simplify the integration of rational functions. Rational functions, which are ratios of two polynomials, can often be challenging to integrate directly. However, by decomposing the rational function into simpler fractions, each of which can be integrated more easily, we can solve the integral more efficiently. This technique is particularly useful when the denominator of the rational function can be factored into linear and/or irreducible quadratic factors. The underlying principle is to express the original rational function as a sum of simpler fractions, where each fraction has a denominator that is one of the factors of the original denominator. The numerators of these simpler fractions are constants or polynomials of a lower degree than the corresponding factors in the denominator. By finding these numerators, we can rewrite the integral as a sum of integrals, each of which is typically much easier to solve. In the context of the given integral, ${ f(x) = \int \frac{1}{(x+5)^2(x-1)} dx }$ , the denominator is already factored, which makes it an ideal candidate for partial fraction decomposition. The factors are ${ (x+5)^2 }$ and ${ (x-1) }$ , indicating that we will need to set up the decomposition with terms corresponding to each of these factors. The success of this method hinges on correctly identifying the factors in the denominator and setting up the appropriate form of the partial fractions. Once the decomposition is set up, the next step involves solving for the unknown coefficients in the numerators of the partial fractions, which typically involves algebraic manipulation and solving a system of equations. With the coefficients determined, the original integral can be rewritten as a sum of simpler integrals, each of which can be evaluated using standard integration techniques.

To effectively apply the method of partial fractions, it's crucial to grasp the underlying principles and steps involved. The technique allows us to break down complex rational functions into simpler components that are easier to integrate. A rational function is essentially a fraction where both the numerator and the denominator are polynomials. When faced with the integral of a rational function, especially one with a denominator that can be factored, partial fraction decomposition offers a systematic way to simplify the integration process. The first step is to ensure that the degree of the numerator is less than the degree of the denominator; if not, polynomial long division must be performed first. Next, factor the denominator into linear and/or irreducible quadratic factors. For each distinct linear factor ${ (ax + b) }$ , include a term of the form ${ \frac{A}{ax + b} }$ in the decomposition. For each repeated linear factor ${ (ax + b)^n }$ , include terms of the form ${ \frac{A_1}{ax + b} + \frac{A_2}{(ax + b)^2} + \cdots + \frac{A_n}{(ax + b)^n} }$ . For each irreducible quadratic factor ${ (ax^2 + bx + c) }$ , include a term of the form ${ \frac{Bx + C}{ax^2 + bx + c} }$ . For each repeated irreducible quadratic factor ${ (ax^2 + bx + c)^n }$ , include terms of the form ${ \frac{B_1x + C_1}{ax^2 + bx + c} + \frac{B_2x + C_2}{(ax^2 + bx + c)^2} + \cdots + \frac{B_nx + C_n}{(ax^2 + bx + c)^n} }$ . Once the decomposition is set up, the next step is to solve for the unknown constants in the numerators. This is typically done by multiplying both sides of the equation by the original denominator, which clears the fractions. Then, by either substituting specific values of ${ x }$ or equating coefficients of like terms, a system of equations can be formed and solved for the unknown constants. With the constants determined, the original integral can be rewritten as a sum of simpler integrals, each of which can be evaluated using standard integration techniques. In the case of our given integral, ${ f(x) = \int \frac{1}{(x+5)^2(x-1)} dx }$ , the denominator ${ (x+5)^2(x-1) }$ factors into a repeated linear factor ${ (x+5)^2 }$ and a distinct linear factor ${ (x-1) }$ , guiding the structure of our partial fraction decomposition.

The core of the problem lies in decomposing the rational function ${ \frac{1}{(x+5)^2(x-1)} }$ into simpler fractions. The denominator consists of a repeated linear factor, ${ (x+5)^2 }$ , and a distinct linear factor, ${ (x-1) }$ . This composition dictates the form of the partial fraction decomposition. We must account for both the repeated factor and the distinct factor to accurately represent the original rational function as a sum of simpler fractions. For the repeated linear factor ${ (x+5)^2 }$ , we will need two terms: one with ${ (x+5) }$ in the denominator and another with ${ (x+5)^2 }$ in the denominator. This ensures that we capture all possible contributions from the repeated factor. For the distinct linear factor ${ (x-1) }$ , we will need a single term with ${ (x-1) }$ in the denominator. The numerators of these fractions will be constants, as the denominators are linear or squared linear terms. Setting up the decomposition involves introducing unknown constants in the numerators of the partial fractions. These constants will need to be determined by solving a system of equations, which arises from equating the numerators after clearing the denominators. The correct setup of the decomposition is crucial because it directly affects the subsequent steps of solving for the constants and ultimately integrating the function. An incorrect setup will lead to an incorrect decomposition and an incorrect integral. Therefore, careful attention must be paid to the form of the factors in the denominator to ensure the partial fraction decomposition is structured appropriately.

To begin the partial fraction decomposition, we express the given rational function as a sum of simpler fractions. Since we have a repeated linear factor ${ (x+5)^2 }$ and a distinct linear factor ${ (x-1) }$ in the denominator, the decomposition will take the form:

${ \frac{1}{(x+5)^2(x-1)} = \frac{A}{x+5} + \frac{B}{(x+5)^2} + \frac{C}{x-1} }$

Here, ${ A }$ , ${ B }$ , and ${ C }$ are constants that we need to determine. Each term on the right-hand side corresponds to one of the factors in the original denominator. The repeated linear factor ${ (x+5)^2 }$ requires two terms: one with ${ (x+5) }$ in the denominator and another with ${ (x+5)^2 }$ in the denominator. The distinct linear factor ${ (x-1) }$ requires a single term with ${ (x-1) }$ in the denominator. The numerators are constants because the denominators are linear or squared linear terms. To find the values of ${ A }$ , ${ B }$ , and ${ C }$ , we will multiply both sides of the equation by the original denominator ${ (x+5)^2(x-1) }$ , which will clear the fractions. This will result in a polynomial equation that we can solve by either substituting specific values of ${ x }$ or equating the coefficients of like terms. The setup of this decomposition is critical because it dictates the form of the simpler fractions that we will integrate. A correct setup ensures that we can accurately represent the original rational function as a sum of manageable terms. Once the constants are determined, we will be able to rewrite the original integral as a sum of simpler integrals, each of which can be evaluated using standard integration techniques. This decomposition is a key step in simplifying the integration process and making the integral solvable.

After setting up the partial fraction decomposition, the next crucial step involves solving for the unknown constants: ${ A }$ , ${ B }$ , and ${ C }$ . To do this, we first multiply both sides of the equation by the original denominator, ${ (x+5)^2(x-1) }$ , to clear the fractions. This gives us:

${ 1 = A(x+5)(x-1) + B(x-1) + C(x+5)^2 }$

This equation is now free of fractions, making it easier to work with. We have a couple of methods to solve for the constants: substituting specific values of ${ x }$ or equating coefficients of like terms. Substituting specific values of ${ x }$ can be particularly effective when we choose values that make certain terms zero, which simplifies the equation. For example, if we let ${ x = 1 }$ , the terms with ${ A }$ and ${ B }$ will become zero, allowing us to solve directly for ${ C }$ . Similarly, if we let ${ x = -5 }$ , the terms with ${ A }$ and ${ C }$ will become zero, allowing us to solve directly for ${ B }$ . Once we have found ${ B }$ and ${ C }$ , we can substitute another value of ${ x }$ (such as ${ x = 0 }$ ) to solve for ${ A }$ . Alternatively, we can expand the right-hand side of the equation and equate the coefficients of like terms. This involves expanding the polynomial expressions and then grouping terms with the same powers of ${ x }$ . By equating the coefficients of ${ x^2 }$ , ${ x }$ , and the constant term on both sides of the equation, we can create a system of linear equations. This system can then be solved using methods such as substitution, elimination, or matrix methods. Regardless of the method used, the goal is to find the values of ${ A }$ , ${ B }$ , and ${ C }$ that satisfy the equation. These constants are essential for completing the partial fraction decomposition and rewriting the original integral in a form that is easier to integrate. Once the constants are known, we can substitute them back into the partial fraction decomposition and proceed with integrating each term separately.

Factors of Each Partial Fraction

Based on the decomposition setup, the factors of each partial fraction are:

For the term ${ \frac{A}{x+5} }$ , the factor is ${ (x+5) }$ .
For the term ${ \frac{B}{(x+5)^2} }$ , the factor is ${ (x+5)^2 }$ .
For the term ${ \frac{C}{x-1} }$ , the factor is ${ (x-1) }$ .

These factors correspond to the denominators of the partial fractions. Understanding these factors is essential for setting up the correct form of the decomposition and for the subsequent integration steps. Each factor represents a part of the original denominator, and the partial fractions allow us to address each part separately. The linear factors ${ (x+5) }$ and ${ (x-1) }$ indicate simple integrals that can be solved using basic integration techniques. The repeated linear factor ${ (x+5)^2 }$ also leads to a straightforward integral, though it requires a slightly different approach than a simple linear factor. By identifying these factors, we can see how the partial fraction decomposition simplifies the integration process by breaking down a complex rational function into more manageable parts. The constants ${ A }$ , ${ B }$ , and ${ C }$ will determine the weight of each of these factors in the overall integral, but the factors themselves define the form of the simpler integrals that we will need to evaluate. This decomposition is a powerful tool for handling rational functions in calculus, making integrals that would otherwise be difficult or impossible to solve much more tractable.

In conclusion, the method of partial fraction decomposition is a vital technique for integrating rational functions. By breaking down a complex fraction into simpler fractions, we can evaluate integrals that would otherwise be challenging to solve. In the case of the integral ${ f(x) = \int \frac{1}{(x+5)^2(x-1)} dx }$ , we successfully decomposed the rational function into partial fractions with factors ${ (x+5) }$ , ${ (x+5)^2 }$ , and ${ (x-1) }$ . This decomposition allows us to rewrite the original integral as a sum of simpler integrals, each of which can be evaluated using standard integration techniques. Understanding and applying partial fraction decomposition is essential for mastering integral calculus and tackling a wide range of integration problems. This technique not only simplifies the integration process but also provides a systematic approach to solving integrals involving rational functions. By identifying the factors in the denominator and setting up the appropriate form of the partial fractions, we can efficiently find the integral and solve complex calculus problems. The decomposition of the rational function into simpler parts makes the integration process more manageable and highlights the power of algebraic manipulation in solving calculus problems.