Chapter 2: Problem 33
Use the method of partial fractions to evaluate each of the following integrals. \(\int \frac{d x}{x^{2}-5 x+6}\)
Short Answer
Expert verified
\( \ln|x-2| - \ln|x-3| + C \)
Step by step solution
01
Factor the Denominator
To use partial fraction decomposition, you first need to factor the denominator. The expression in the denominator is a quadratic: \(x^2 - 5x + 6\). Try to find two numbers that multiply to 6 and add to -5. These numbers are -2 and -3, so the factorization is: \[(x^2 - 5x + 6) = (x-2)(x-3)\]
02
Set Up Partial Fractions
Express the integrand \(\frac{1}{(x-2)(x-3)}\) as a sum of partial fractions. Assume it can be written in the form: \[\frac{A}{x-2} + \frac{B}{x-3}\] where \(A\) and \(B\) are constants to be determined.
03
Solve for Constants A and B
To find the values of \(A\) and \(B\), equate the expressions: \[\frac{1}{(x-2)(x-3)} = \frac{A}{x-2} + \frac{B}{x-3}\] Clear the denominators by multiplying through by \((x-2)(x-3)\): \[1 = A(x-3) + B(x-2)\] Expand and collect like terms: \[1 = A x - 3A + B x - 2B = (A + B)x - (3A + 2B)\] Equating coefficients gives two equations: \[A + B = 0\] and \[-3A - 2B = 1\] Solve these equations to find \(A = 1\) and \(B = -1\).
04
Write the Partial Fraction Decomposition
Substitute the values of \(A\) and \(B\) back into the partial fraction decomposition: \[\frac{1}{(x-2)(x-3)} = \frac{1}{x-2} - \frac{1}{x-3}\]
05
Integrate the Partial Fractions
Integrate each term separately: \[ \int \left( \frac{1}{x-2} - \frac{1}{x-3} \right) \, dx = \int \frac{1}{x-2} \, dx - \int \frac{1}{x-3} \, dx\] This yields: \[ \ln|x-2| - \ln|x-3| + C \] where \(C\) is the constant of integration.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Integration
Integration is a fundamental concept in calculus that involves finding the antiderivative of a function. It's like the reverse process of differentiation, where you find the original function from its derivative.
In this exercise, we use integration to solve \[ \int \frac{dx}{x^2 - 5x + 6} \]by expressing it in a simpler form using partial fractions. The goal is to break down the complex rational expression into more manageable pieces. After successfully doing this, each part can be integrated separately. Through this decomposition, we are finding simpler functions whose derivatives add up to the integrand, allowing us to evaluate the integral more easily.
It’s important to ensure that each step of integration is carefully handled, from setting up the integral with partial fractions to solving it with basic logarithmic rules. The result of the integration gives us the antiderivative of the original expression, plus a constant of integration, often denoted as \( C \).
In this exercise, we use integration to solve \[ \int \frac{dx}{x^2 - 5x + 6} \]by expressing it in a simpler form using partial fractions. The goal is to break down the complex rational expression into more manageable pieces. After successfully doing this, each part can be integrated separately. Through this decomposition, we are finding simpler functions whose derivatives add up to the integrand, allowing us to evaluate the integral more easily.
It’s important to ensure that each step of integration is carefully handled, from setting up the integral with partial fractions to solving it with basic logarithmic rules. The result of the integration gives us the antiderivative of the original expression, plus a constant of integration, often denoted as \( C \).
Algebraic Factorization
Algebraic factorization is an essential step when working on integration problems involving rational expressions. It simplifies the given expression and prepares it for partial fraction decomposition.
In this problem, you need to factorize the quadratic expression \(x^2 - 5x + 6\). You look for two numbers that multiply to the constant term (6) and add to the linear coefficient (-5). These numbers are \(-2\) and \(-3\), resulting in the factorization:
\[(x-2)(x-3)\]
This step not only simplifies the expression but reveals the individual linear factors. These provide the necessary foundation for expressing the rational expression as a sum of simpler fractions. Once factorized, each part of the expression can be addressed independently, facilitating the integration process by splitting the original complex expression into simpler, more approachable components.
In this problem, you need to factorize the quadratic expression \(x^2 - 5x + 6\). You look for two numbers that multiply to the constant term (6) and add to the linear coefficient (-5). These numbers are \(-2\) and \(-3\), resulting in the factorization:
\[(x-2)(x-3)\]
This step not only simplifies the expression but reveals the individual linear factors. These provide the necessary foundation for expressing the rational expression as a sum of simpler fractions. Once factorized, each part of the expression can be addressed independently, facilitating the integration process by splitting the original complex expression into simpler, more approachable components.
Rational Expressions
Rational expressions are fractions where the numerator and denominator are polynomial expressions. They can present challenges in calculus, particularly when integrating.
In the exercise, the integrand is a rational expression \( \frac{1}{x^2 - 5x + 6} \). Dealing with such expressions often requires breaking them down into smaller, simpler fractions through a process called partial fraction decomposition.
This method relies on firstly factorizing the denominator. The factorized form allows us to express the rational expression as a sum of simpler fractions, like \( \frac{1}{x-2} \) and \( \frac{-1}{x-3} \).
These simpler forms are easier to integrate, simplifying the process significantly. By converting a difficult integral into a set of simpler integrals, you effectively tackle complex rational expressions with more clarity and less effort. This approach is particularly powerful in calculus when dealing with polynomial-based fractions.
In the exercise, the integrand is a rational expression \( \frac{1}{x^2 - 5x + 6} \). Dealing with such expressions often requires breaking them down into smaller, simpler fractions through a process called partial fraction decomposition.
This method relies on firstly factorizing the denominator. The factorized form allows us to express the rational expression as a sum of simpler fractions, like \( \frac{1}{x-2} \) and \( \frac{-1}{x-3} \).
These simpler forms are easier to integrate, simplifying the process significantly. By converting a difficult integral into a set of simpler integrals, you effectively tackle complex rational expressions with more clarity and less effort. This approach is particularly powerful in calculus when dealing with polynomial-based fractions.
Calculus
Calculus is the branch of mathematics that studies continuous change. It is widely used in various scientific fields, such as physics and engineering.
The exercise provided involves using calculus, specifically integration, to find the antiderivative of a rational expression. Partial fraction decomposition is a technique that is often leveraged in calculus to simplify integrals involving rational expressions.
Within calculus, integrating a rational expression requires understanding both algebraic manipulation and the fundamentals of integration. By using partial fractions, the integrals of simple fractions can be found using basic calculus rules, like the natural logarithm. This approach reduces complexity, providing clear steps to solve integrals that otherwise might seem daunting.
Understanding how to switch between algebraic manipulation and integral calculation is key in mastering problems involving calculus. It fine-tunes both analytical and problem-solving skills by turning complex mathematical problems into manageable solutions.
The exercise provided involves using calculus, specifically integration, to find the antiderivative of a rational expression. Partial fraction decomposition is a technique that is often leveraged in calculus to simplify integrals involving rational expressions.
Within calculus, integrating a rational expression requires understanding both algebraic manipulation and the fundamentals of integration. By using partial fractions, the integrals of simple fractions can be found using basic calculus rules, like the natural logarithm. This approach reduces complexity, providing clear steps to solve integrals that otherwise might seem daunting.
Understanding how to switch between algebraic manipulation and integral calculation is key in mastering problems involving calculus. It fine-tunes both analytical and problem-solving skills by turning complex mathematical problems into manageable solutions.
