Problem 19
Question
Write out the partial-fraction decomposition of the function \(f(x)\). $$ f(x)=\frac{x+1}{x(x+2)} $$
Step-by-Step Solution
Verified Answer
The partial-fraction decomposition of \( f(x) = \frac{x+1}{x(x+2)} \) is \( \frac{1/2}{x} + \frac{1/2}{x+2} \).
1Step 1: Identify Denominator for Decomposition
The given function is \( f(x) = \frac{x+1}{x(x+2)} \). The denominator is \( x(x+2) \), which can be decomposed into the linear factors \( x \) and \( x+2 \).
2Step 2: Setup Partial Fractions
Because the denominator consists of two linear factors, express \( \frac{x+1}{x(x+2)} \) as a sum of two fractions: \( \frac{A}{x} + \frac{B}{x+2} \).
3Step 3: Clear the Denominator
Multiply both sides of the equation by \( x(x+2) \) to eliminate the denominators: \( x+1 = A(x+2) + Bx \).
4Step 4: Expand and Collect Like Terms
Expand the right side of the equation: \( x+1 = Ax + 2A + Bx \).Combine like terms to get: \( x+1 = (A+B)x + 2A \).
5Step 5: Solve for Coefficients
Equate the coefficients of like powers of \( x \) from both sides of the equation:For \( x \): \( A+B = 1 \)Constant term: \( 2A = 1 \).
6Step 6: Calculate the Values of A and B
From \( 2A = 1 \), we find \( A = \frac{1}{2} \).Substitute \( A = \frac{1}{2} \) into \( A+B = 1 \), giving \( \frac{1}{2} + B = 1 \), which leads to \( B = \frac{1}{2} \).
7Step 7: Write the Partial Fraction Decomposition
Substitute the values of \( A \) and \( B \) back into the partial fraction setup: \( \frac{x+1}{x(x+2)} = \frac{1/2}{x} + \frac{1/2}{x+2} \).
Key Concepts
CalculusRational FunctionsAlgebraic Manipulation
Calculus
Partial fraction decomposition is a significant topic in calculus, especially when dealing with integrations. It becomes handy when integrating rational functions – functions that can be expressed as one polynomial divided by another. Knowing how to decompose a function is crucial when the integral of a single rational function is complex and hard to solve directly.
When you decompose a fraction into partial fractions, you express it as a sum of simpler fractions. These simpler fractions are often easier to integrate directly, using basic integration rules for logarithms or algebraic terms. This technique is particularly useful when tackling integrals involving rational functions, where antiderivatives may not be obvious or easy.
Not only do partial fractions simplify integrations, but they also help in solving differential equations. Through calculus, mastering partial fraction decomposition gives a robust toolkit for dealing with complex mathematical models and provides deeper insights into the behavior of functions within these systems.
When you decompose a fraction into partial fractions, you express it as a sum of simpler fractions. These simpler fractions are often easier to integrate directly, using basic integration rules for logarithms or algebraic terms. This technique is particularly useful when tackling integrals involving rational functions, where antiderivatives may not be obvious or easy.
Not only do partial fractions simplify integrations, but they also help in solving differential equations. Through calculus, mastering partial fraction decomposition gives a robust toolkit for dealing with complex mathematical models and provides deeper insights into the behavior of functions within these systems.
Rational Functions
Rational functions are a fundamental concept in both algebra and calculus. These functions are defined as the quotient of two polynomials. The general form is \( f(x) = \frac{p(x)}{q(x)} \), where \( p(x) \) and \( q(x) \) are polynomials and \( q(x) eq 0 \).
Understanding rational functions includes recognizing their domain, which is all real numbers except where the denominator is zero. This awareness helps us in identifying the points where the function is undefined or where it may have vertical asymptotes.
In the context of partial fraction decomposition, rational functions are taken apart into simpler, more manageable fractions. These resulting fractions retain the properties of rational functions but are easier to work with. Decomposing rationals is like breaking down a complex task into smaller, more approachable parts, thus making subsequent operations, like integration or solving, more straightforward.
Understanding rational functions includes recognizing their domain, which is all real numbers except where the denominator is zero. This awareness helps us in identifying the points where the function is undefined or where it may have vertical asymptotes.
In the context of partial fraction decomposition, rational functions are taken apart into simpler, more manageable fractions. These resulting fractions retain the properties of rational functions but are easier to work with. Decomposing rationals is like breaking down a complex task into smaller, more approachable parts, thus making subsequent operations, like integration or solving, more straightforward.
Algebraic Manipulation
Algebraic manipulation is essential for handling rational functions, particularly in partial fraction decomposition. It involves rearranging and simplifying expressions to find solutions or understand properties of mathematical statements.
In partial fraction decomposition, algebraic manipulation allows us to express a single complex rational function as the sum of simpler ones. Here's a step-by-step look at such manipulations using the given problem:
In partial fraction decomposition, algebraic manipulation allows us to express a single complex rational function as the sum of simpler ones. Here's a step-by-step look at such manipulations using the given problem:
- First, the denominator is broken down into linear factors, \( x \) and \( x+2 \).
- Then, set up an equation with constants \( A \) and \( B \) representing the numerators of these simpler fractions: \( \frac{A}{x} + \frac{B}{x+2} \).
- Next, eliminate the denominators by multiplying through by the original denominator, \( x(x+2) \), to simplify.
- Expand the results and collect terms to relate this expression back to the original function.
- Finally, solve for \( A \) and \( B \) using algebraic equations derived from matching coefficients.
Other exercises in this chapter
Problem 19
Determine whether each integral is convergent. If the integral is convergent, compute its value. $$ \int_{1}^{\infty} \frac{1}{x^{3}} d x $$
View solution Problem 19
In Problems 1-30, use integration by parts to evaluate each integral. $$ \int_{1}^{2} \ln x d x $$
View solution Problem 19
In Problems 17-36, use substitution to evaluate each indefinite integral. $$ \int(4 x-1) \sqrt{2 x^{2}-x+2} d x $$
View solution Problem 20
Use the Table of Integrals to compute each integral after manipulating the integrand in a suitable way. $$ \int_{1}^{e}(x+2)^{2} \ln x d x $$
View solution