Problem 15
Question
In Problems , use partial-fraction decomposition to evaluate the integrals. $$ \int \frac{1}{(x+1)(x-3)} d x $$
Step-by-Step Solution
Verified Answer
\(-\frac{1}{4} \ln|x+1| + \frac{1}{4} \ln|x-3| + C\).
1Step 1: Set Up Partial-Fraction Decomposition
The integrand has the form \( \frac{1}{(x+1)(x-3)} \). We express it as the sum of partial fractions: \( \frac{A}{x+1} + \frac{B}{x-3} \).
2Step 2: Clear the Denominator
Multiply the entire equation by \((x+1)(x-3)\) to clear the fraction: \(1=A(x-3) + B(x+1)\).
3Step 3: Expand and Collect Like Terms
Expand the right side: \(1 = A(x-3) + B(x+1) = Ax - 3A + Bx + B\). Combine like terms to get: \(1 = (A+B)x + (-3A+B)\).
4Step 4: Match Coefficients
Match the coefficients from both sides where \(x\) terms must match and constant terms must match: \(A+B=0\) and \(-3A+B=1\).
5Step 5: Solve for Constants A and B
Solving the equations \(A+B=0\) and \(-3A+B=1\), we find \(A=-\frac{1}{4}\) and \(B=\frac{1}{4}\).
6Step 6: Integrate the Partial Fractions
Now, rewrite the integral using these constants: \( \int \left(\frac{-1/4}{x+1} + \frac{1/4}{x-3}\right) dx \). This simplifies to \( -\frac{1}{4} \int \frac{1}{x+1} \, dx + \frac{1}{4} \int \frac{1}{x-3} \, dx \).
7Step 7: Evaluate Each Integral
The integrals are standard logarithmic forms: \(-\frac{1}{4} \ln|x+1| + \frac{1}{4} \ln|x-3| + C\), where \(C\) is the constant of integration.
Key Concepts
Understanding IntegralsLogarithmic IntegrationApplying Algebraic Equations
Understanding Integrals
Integrals are a fundamental part of calculus, acting as the inverse operation to differentiation. When solving an integral, we often attempt to find the original function that, when differentiated, results in the given integrand. For example, if we say that the derivative of a function is a specific result, the integral helps us reverse this operation and find the original function.
In our exercise, we come across an integral of the form \( \int \frac{1}{(x+1)(x-3)} \, dx \). This is a rational function where integration requires specific techniques, such as partial fraction decomposition, to simplify the integrand so that standard integration techniques can be applied effectively.
To tackle such problems, it is crucial to understand how to manipulate the integrand into a more workable form. This manipulation turns complex fractions into the sum of simpler fractions, allowing us to use basic integration rules. It transforms the task into a series of smaller, more manageable integrals.
In our exercise, we come across an integral of the form \( \int \frac{1}{(x+1)(x-3)} \, dx \). This is a rational function where integration requires specific techniques, such as partial fraction decomposition, to simplify the integrand so that standard integration techniques can be applied effectively.
To tackle such problems, it is crucial to understand how to manipulate the integrand into a more workable form. This manipulation turns complex fractions into the sum of simpler fractions, allowing us to use basic integration rules. It transforms the task into a series of smaller, more manageable integrals.
Logarithmic Integration
Logarithmic integration is a technique used when integrating functions of the form \( \frac{1}{x+a} \). The integral of such a function is logarithmic and given by \( \int \frac{1}{x+a} \, dx = \ln|x+a| + C \), where \(C\) is the constant of integration.
In our solution approach, we notice this form appearing after the partial fraction decomposition of the original integrand. By decomposing the function \( \frac{1}{(x+1)(x-3)} \) into simpler parts like \( \frac{A}{x+1} + \frac{B}{x-3} \), each piece can be integrated separately using the logarithmic integration technique.
When each fraction involves a simple linear denominator such as \(x+1\) or \(x-3\), this method yields natural logarithms, simplifying the process. The resulting expression from integrating these logs helps reach the solution form, efficiently solving the integral step by step.
In our solution approach, we notice this form appearing after the partial fraction decomposition of the original integrand. By decomposing the function \( \frac{1}{(x+1)(x-3)} \) into simpler parts like \( \frac{A}{x+1} + \frac{B}{x-3} \), each piece can be integrated separately using the logarithmic integration technique.
When each fraction involves a simple linear denominator such as \(x+1\) or \(x-3\), this method yields natural logarithms, simplifying the process. The resulting expression from integrating these logs helps reach the solution form, efficiently solving the integral step by step.
Applying Algebraic Equations
Algebraic equations play a crucial role in setting up and solving the partial fraction decomposition. During the problem-solving process, once we express the complex fraction in terms of simpler fractions \( \frac{A}{x+1} + \frac{B}{x-3} \), we encounter the necessity to determine the values of constants \(A\) and \(B\).
Clearing the denominators by multiplying through by \((x+1)(x-3)\) turns the equation into a form where we can equate coefficients of powers of \(x\). This allows us to solve algebraic equations such as \(A + B = 0\) and \(-3A + B = 1\) to find \(A = -\frac{1}{4}\) and \(B = \frac{1}{4}\).
Clearing the denominators by multiplying through by \((x+1)(x-3)\) turns the equation into a form where we can equate coefficients of powers of \(x\). This allows us to solve algebraic equations such as \(A + B = 0\) and \(-3A + B = 1\) to find \(A = -\frac{1}{4}\) and \(B = \frac{1}{4}\).
- This step involves several methods, such as substitution or elimination, to find the values of unknowns.
- These constants then plug back into the decomposed fraction to allow for individual integration.
Other exercises in this chapter
Problem 15
In Problems 11-16, compute the Taylor polynomial of degree \(n\) about \(a=0\) for the indicated functions and compare the value of the functions at the indicat
View solution Problem 15
All the integrals are improper and converge. Explain in each case why the integral is improper, andevaluate each integral. $$ \int_{-1}^{1} \ln |x| d x $$
View solution Problem 16
Use integration by parts to evaluate the integrals. $$ \int x \csc ^{2} x d x $$
View solution Problem 16
Use the trapezoidal rule to approximate each integral with the specified value of \(n .\) Compare your approximation with the exact value. \(\int_{1}^{2} \frac{
View solution