Problem 26 - Thomas Calculus

Question

In Exercises \(21-28,\) express the integrands as a sum of partial fractions and evaluate the integrals. $$ \int \frac{s^{4}+81}{s\left(s^{2}+9\right)^{2}} d s $$

Step-by-Step Solution

Answer

The integral evaluates to \( \ln|s| + C \).

Step 1: 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: \( \frac{A}{s} + \frac{Bs+C}{s^2+9} + \frac{Ds+E}{(s^2+9)^2} \).

Step 2: 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 \(E\).

Step 3: 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:- \(A = 1\),- \(B = 0\),- \(C = 0\),- \(D = 0\),- \(E = 0\).

Step 4: 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.

Step 5: Simplify and Provide Final Answer

The integral evaluates to:\[ \int \frac{1}{s} ds = \ln|s| + C, \]where \( C \) is the constant of integration.

Key Concepts

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.

Breaking down these complex expressions into simpler fractions is essential for solving integrals, as it allows us to handle each part separately. This technique becomes especially useful when dealing with higher degree polynomials in calculus.

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 \( B = C = D = E = 0 \).

This process of identifying coefficients involves creating an equation by multiplying through by the common denominator, expanding, and then equating terms. Each coefficient plays a crucial role in accurately breaking down the rational expression and thus simplifying the integration.

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.

By focusing on each integral individually, we follow a methodical approach that simplifies the overall process. This technique not only clarifies the solution steps but also ensures no integral component is overlooked. Combining the results of these smaller integrations provides a comprehensive answer to the original problem.