Chapter 8: Problem 56
Use any method to evaluate the integrals in Exercises \(55-66\) $$ \int \frac{x+2}{x^{3}-2 x^{2}-3 x} d x $$
Short Answer
Expert verified
Factor the integrand, decompose into partial fractions, and integrate each term separately.
Step by step solution
01
Factor the Denominator
First, factor the denominator \(x^3 - 2x^2 - 3x\). Notice that \(x\) is a common factor, so we factor it out: \(x(x^2 - 2x - 3)\). To factor \(x^2 - 2x - 3\), find two numbers whose product is \(-3\) and sum is \(-2\). These numbers are \(-3\) and \(1\), so \(x^2 - 2x - 3 = (x - 3)(x + 1)\). Thus, \(x^3 - 2x^2 - 3x = x(x - 3)(x + 1)\).
02
Set Up Partial Fraction Decomposition
Express the given integrand using partial fraction decomposition: \(\frac{x+2}{x(x-3)(x+1)} = \frac{A}{x} + \frac{B}{x-3} + \frac{C}{x+1}\). This will allow us to evaluate the integral by splitting it into simpler parts.
03
Solve for Coefficients
Multiply through by the original denominator \(x(x-3)(x+1)\) to clear fractions: \((x+2) = A(x-3)(x+1) + Bx(x+1) + Cx(x-3)\). Expand and collect like terms to form an equation. By substituting suitable values for \(x\) (e.g., \(x = 0, 3, -1\)) and using algebraic techniques, solve for \(A\), \(B\), and \(C\).
04
Integrate Each Term
Once the coefficients \(A\), \(B\), and \(C\) are found, integrate each term: \(\int \frac{A}{x} \, dx + \int \frac{B}{x-3} \, dx + \int \frac{C}{x+1} \, dx\). These integrals yield: \(A \ln|x| + B \ln|x-3| + C \ln|x+1| + C\).
05
Combine Results
Combine the results from the integration in Step 4 to write the complete solution for the original integral. Simplify if necessary.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Partial Fraction Decomposition
Partial Fraction Decomposition is a technique used to express a complex rational expression as a sum of simpler fractions. This helps in simplifying the process of integration, especially when dealing with polynomial denominators. In the given exercise, the fraction \( \frac{x+2}{x(x-3)(x+1)} \) is decomposed into partial fractions in order to separate it into more manageable parts. This fraction can be represented as \( \frac{A}{x} + \frac{B}{x-3} + \frac{C}{x+1} \).
To determine the coefficients \(A\), \(B\), and \(C\), we multiply through by the denominator \(x(x-3)(x+1)\), set the resulting expression equal to \(x+2\), and solve for the unknowns by substituting suitable values for \(x\). This method simplifies the integration process by breaking down a complex problem into parts that can be integrated using basic techniques.
To determine the coefficients \(A\), \(B\), and \(C\), we multiply through by the denominator \(x(x-3)(x+1)\), set the resulting expression equal to \(x+2\), and solve for the unknowns by substituting suitable values for \(x\). This method simplifies the integration process by breaking down a complex problem into parts that can be integrated using basic techniques.
Factoring Polynomials
Factoring polynomials is a crucial step when working with polynomial expressions, particularly for simplifying fractions in partial fraction decomposition.
This involves breaking down a polynomial into its constituent factors, which are simpler polynomials that, when multiplied together, give the original polynomial. In this exercise, the polynomial denominator \( x^3 - 2x^2 - 3x \) was factored by first extracting the common factor \( x \), leading to \( x(x^2 - 2x - 3) \).
This involves breaking down a polynomial into its constituent factors, which are simpler polynomials that, when multiplied together, give the original polynomial. In this exercise, the polynomial denominator \( x^3 - 2x^2 - 3x \) was factored by first extracting the common factor \( x \), leading to \( x(x^2 - 2x - 3) \).
- The quadratic \(x^2 - 2x - 3\) is then further factored into \((x-3)(x+1)\) by finding numbers that multiply to \(-3\) and add to \(-2\).
Integrals
Integrals are fundamental to calculus and represent the area under a curve or the accumulated quantity of a function over an interval. By applying polynomial factoring and partial fraction decomposition, the task of integrating a complex rational function becomes manageable.
- Once the expression is decomposed, each partial fraction is integrated individually.
- In this case, each fraction leads to a natural logarithm function due to the linear denominators.
Logarithmic Integration
Logarithmic Integration comes into play when integrating fractions with a linear denominator, resulting in expressions that involve logarithms. After decomposing the original problem into partial fractions, each term such as \( \int \frac{A}{x} \, dx \) can be integrated using the natural logarithm rule.
For example, integrating \( \frac{A}{x} \) results in \( A\ln|x| \), and similar for other terms involving \(x-3\) or \(x+1\).
For example, integrating \( \frac{A}{x} \) results in \( A\ln|x| \), and similar for other terms involving \(x-3\) or \(x+1\).
- In each case, the integration of \( \frac{1}{u} \, du \) yields \( \ln|u| + C \), where \(u\) represents the linear expression in the denominator.
- This results in a straightforward solution formed by the sum of these logarithm terms.