Chapter 8: Problem 26
In Exercises \(21-28,\) express the integrands as a sum of partial fractions and evaluate the integrals. $$ \int \frac{s^{4}+81}{s\left(s^{2}+9\right)^{2}} d s $$
Short Answer
Expert verified
The integral evaluates to \( \ln|s| + C \).
Step by step solution
01
Identify the Partial Fraction Form
The rational expression can be broken down using partial fractions. For the expression \( \frac{s^4 + 81}{s(s^2 + 9)^2} \), start by assuming a decomposition of the form: \( \frac{A}{s} + \frac{Bs+C}{s^2+9} + \frac{Ds+E}{(s^2+9)^2} \).
02
Solve for Coefficients
Multiply both sides by the common denominator \( s(s^2+9)^2 \) to clear the fractions: \[ s^4 + 81 = A(s^2 + 9)^2 + (Bs + C)s(s^2 + 9) + (Ds + E)s. \]Expand and equate the coefficients of like terms to find \(A, B, C, D,\) and \(E\).
03
Match Coefficients to Find Values
After expanding, you will have:\[ A(s^4 + 18s^2 + 81) + (Bs + C)(s^3 + 9s) + (Ds + E)s = s^4 + 81. \]Compare coefficients to find:- \(A = 1\),- \(B = 0\),- \(C = 0\),- \(D = 0\),- \(E = 0\).
04
Integrate Each Term
Integrate each term separately: - \( \int \frac{1}{s} ds, \) which gives \( \ln|s| \),- The terms with \( B, C, D, E \) are zero, so they simplify out.Add the result for the complete integral.
05
Simplify and Provide Final Answer
The integral evaluates to:\[ \int \frac{1}{s} ds = \ln|s| + C, \]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.
Rational Expressions
A rational expression is similar to a fraction, but instead of having just numbers in its numerator and denominator, it involves polynomials. In simpler terms, it's a polynomial divided by another polynomial. Understanding rational expressions is crucial for dealing with partial fraction decomposition, which is a technique used in integration.
- In our exercise, the rational expression is \( \frac{s^4 + 81}{s(s^2 + 9)^2} \).
- The goal is to break it into simpler, more manageable parts.
Integrals
Integrals are a key concept in calculus, often used to find the area under a curve or to accumulate quantities. They enable us to reverse the process of differentiation. In mathematics, integration transforms a function into another function.
Integrals can be particularly tricky when dealing with complex rational expressions. That's where partial fraction decomposition comes in handy. By expressing a complicated fraction as a series of simpler ones, each piece becomes more straightforward to integrate.
In our exercise, once the expression is broken into partial fractions, integrating each term becomes a step-by-step process until the whole expression is simplified. This step-by-step approach enables a systematic solution to otherwise complicated integrals.
Coefficients
To complete a partial fraction decomposition, we need to determine the values of unknown coefficients. These coefficients multiply each part of the decomposition, converting it effectively to equivalent complex expressions.
- In the given example, our decomposition included coefficients \( A, B, C, D, \) and \( E \).
- Through equating and simplifying, we found that \( A = 1 \) and \( B = C = D = E = 0 \).
Step-by-Step Integration
Step-by-step integration allows us to address each component of our decomposed rational expression individually. This method is particularly effective because it breaks down a daunting problem into smaller, manageable tasks.
- First, each term is integrated separately. In our case, \( \int \frac{1}{s} ds \) simplifies to \( \ln|s| \).
- Because coefficients \( B, C, D, \) and \( E \) were found to be zero, the additional terms in the decomposition are ignored.