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Chapter 6: Problem 21

In each of Exercises \(21-26,\) use Heaviside's method to calculate the partial fraction decomposition of the given rational function. $$ \frac{x+1}{(2 x+7)(2 x+9)} $$

Short Answer

Expert verified
The partial fraction decomposition is \( \frac{7}{2x+7} - \frac{5}{2x+9} \).

Step by step solution

01

Express the Rational Function

The given rational function is \( \frac{x+1}{(2x+7)(2x+9)} \). We want to decompose this into partial fractions. We express it as: \( \frac{A}{2x+7} + \frac{B}{2x+9} \), where \( A \) and \( B \) are constants to be determined.
02

Set Up Equation for Heaviside's Method

Multiply the entire equation by \((2x+7)(2x+9)\) to clear the denominators. You obtain: \( x + 1 = A(2x+9) + B(2x+7) \).
03

Solve for A by Heaviside Cover-Up

To find \( A \), substitute \(-\frac{9}{2}\) into the equation since this makes \(2x+9 = 0\). Substituting gives:\( A = \frac{-\frac{9}{2}+1}{-\frac{1}{2}} = \frac{-\frac{7}{2}}{-\frac{1}{2}} = 7 \).
04

Solve for B by Heaviside Cover-Up

For \( B \), substitute \(-\frac{7}{2}\) into the equation to eliminate \( A \), as this makes \(2x+7 = 0\). Substituting gives:\( B = \frac{-\frac{7}{2}+1}{\frac{1}{2}} = \frac{-\frac{5}{2}}{\frac{1}{2}} = -5 \).
05

Write the Partial Fraction Decomposition

Now that we have \( A = 7 \) and \( B = -5 \), the partial fraction decomposition of the rational function is:\( \frac{7}{2x+7} - \frac{5}{2x+9} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Heaviside's Method
Heaviside's Method is a clever and efficient way to perform partial fraction decomposition. It works best with simple linear factors in the denominator. The method greatly simplifies the process of finding constants like \( A \) or \( B \) without tedious algebraic expansion or equating like terms. The basic idea is to isolate each term of the partial fraction by strategically eliminating others.
  • Start by multiplying the entire equation by the common denominator to clear fractions and set up an equation focused only on the numerators.
  • Use the roots of the linear factors to quickly solve for unknown constants. For example, if the denominator includes \(2x + 7\), setting \(2x + 7 = 0\) isolates terms that include its coefficient.
  • This method quickly leads us to each constant by simplifying subsets of the given equation.
Rational Function
A rational function is a quotient of two polynomials. In simpler terms, it's a fraction where the numerator and denominator are both polynomial expressions.
These can often look daunting, with various powers of \( x \), but breaking them down into partial fractions can illuminate their structure and simplify integration or differentiation tasks.
For instance, in our exercise, the rational function \( \frac{x+1}{(2x+7)(2x+9)} \) appeared complex at first glance. However, once decomposed, it simplifies into easy-to-handle fractions: \( \frac{7}{2x+7} \) and \( -\frac{5}{2x+9} \).
  • This simplification is especially useful in calculus, where it helps integrate certain types of functions.
  • The form reveals hidden behaviors in functions that might be obscured by a more complex expression.
  • It offers insights into the function鈥檚 asymptotes and intercepts.
Constants Solution
Determining constants \( A \) and \( B \) in a partial fraction decomposition is crucial for expressing the function as a sum of simpler fractions. With Heaviside鈥檚 Method, once you clear the fractions, finding \( A \) and \( B \) becomes a straightforward process:
  • Solve the resulting equation at the values where other terms drop out due to zeroing the factors.
  • In the exercise, substituting specific roots of the denominator allowed us to solve independently for each constant.
  • Substitute constants back into the partial fraction expression to verify your solution is correct.
  • If the results seem incorrect, it's prudent to check the equation after substituting values, ensuring arithmetic or algebraic errors didn't occur.
Cover-Up Method
The Cover-Up Method is a handy technique for finding constants when dealing with partial fraction decompositions, specifically when all denominators are linear.
Here's a simple way to understand it:
  • You multiply the equation by the common denominator, canceling out fractions, and set each fraction equal by covering or making zero the irrelevant parts.
  • For instance, to find \( A \) in our problem, eliminate the \( B \) term by plugging in the value that zeroes it: \(-\frac{9}{2}\). This immediately isolates terms associated with \( A \).
  • Similarly, cover up the \( A \) term to find \( B \). Here, \(-\frac{7}{2}\) zeros out \(2x + 7\), allowing us to focus on \( B \).
  • This technique is especially valuable because it鈥檚 faster than simultaneously solving linear equations for each constant.

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Most popular questions from this chapter

Use the Comparison Theorem to establish that the given improper integral is divergent. $$ \int_{1}^{\infty} \frac{\exp (x)}{x \exp (x)-1} d x $$

Suppose that \(A\) and \(B\) are constants, not both of which are 0\. Notice that $$ A \cos (x)+B \sin (x)=\sqrt{A^{2}+B^{2}}(\cos (x) \cdot \xi+\sin (x) \cdot \eta) $$ where \(\xi=A / \sqrt{A^{2}+B^{2}}\) and \(\eta=B / \sqrt{A^{2}+B^{2}}\). By observing that \((\xi, \eta)\) is a point on the unit circle, deduce that \(A \cos (x)+\) \(B \sin (x)=\sqrt{A^{2}+B^{2}} \cos (x+\phi)\) for some \(\phi .\) Evaluate $$ \int \frac{1}{(A \cos (x)+B \sin (x))^{k}} d x $$ for \(k=1\) and \(k=2\).

In each of Exercises \(41-54,\) determine whether the given improper integral is convergent or divergent. If it converges, then evaluate it. \(\int_{-1}^{3} \ln (|x|) d x\)

Evaluate the given definite integral. \(\int_{1}^{2} \frac{3 x^{4}+8 x^{2}+3}{x\left(x^{2}+1\right)^{2}} d x\)

For what positive values of \(p\) is the improper integral \(\int_{1}^{e} \frac{1}{x \ln ^{p}(x)} d x\) convergent?
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