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Say which formula, if any, to apply from the table of integrals. Give the values of any constants. $$\int \frac{4 x-2}{x^{2}-9} d x$$

Short Answer

Expert verified
Apply partial fraction decomposition, use \( A = \frac{5}{3} \), \( B = \frac{7}{3} \), integrate to get \( \frac{5}{3} \ln |x-3| + \frac{7}{3} \ln |x+3| + C \).

Step by step solution

01

Identify the Integral Type

The integral to evaluate is \( \int \frac{4x-2}{x^2-9} \, dx \). This resembles a rational function of the form \( \frac{A}{(x^2 - a^2)} \). Recognize that the denominator \( x^2 - 9 \) can be factored as \( (x+3)(x-3) \). This suggests using partial fraction decomposition.
02

Set Up Partial Fraction Decomposition

We express \( \frac{4x-2}{x^2-9} \) as \( \frac{A}{x-3} + \frac{B}{x+3} \). To find constants \( A \) and \( B \), multiply both sides by \( x^2-9 \): \( 4x-2 = A(x+3) + B(x-3) \).
03

Solve for Constants

Expand and equate the equation: \( 4x-2 = Ax + 3A + Bx - 3B \). Combining like terms gives \( (A+B)x + (3A-3B) = 4x-2 \). This implies \( A+B = 4 \) and \( 3A-3B = -2 \). Solve these equations.
04

Step 3.1: Solve System of Equations

From \( A + B = 4 \), we express \( A = 4 - B \). Substituting in the second equation \( 3A - 3B = -2 \), we have \( 3(4 - B) - 3B = -2 \). Solving gives \( 12 - 3B - 3B = -2 \), so \( -6B = -14 \), thus \( B = \frac{7}{3} \). Therefore, \( A = 4 - \frac{7}{3} = \frac{5}{3} \).
05

Substitute Constants into Partial Fractions

Substitute \( A \) and \( B \) back into the partial fractions: \( \frac{4x-2}{x^2-9} = \frac{5/3}{x-3} + \frac{7/3}{x+3} \).
06

Integrate Each Term Separately

Integrate the expression: \( \int \frac{5/3}{x-3} \, dx + \int \frac{7/3}{x+3} \, dx \). For \( \int \frac{k}{x-a} \, dx \), the result is \( k \ln |x-a| + C \). This yields \( \frac{5}{3} \ln |x-3| + \frac{7}{3} \ln |x+3| + C \).

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Key Concepts

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

Partial Fraction Decomposition
Partial fraction decomposition is a technique used in calculus, especially when dealing with integrals of rational functions. It helps break down a complicated fraction into simpler, more manageable pieces.

Here’s how it generally works:
  • First, ensure your rational function is proper, meaning the degree of the numerator is less than the degree of the denominator. If not, perform polynomial long division first.
  • Next, factor the denominator completely. For example, in the function \(\frac{4x-2}{x^2-9}\), the denominator \(x^2-9\) can be factored as \((x+3)(x-3)\).
  • Write the decomposed fraction as a sum of simpler fractions whose denominators are those factors. Assign constants (usually \(A, B, C\), etc.) to the numerators. For instance, \(\frac{4x-2}{(x-3)(x+3)} = \frac{A}{x-3} + \frac{B}{x+3}\).
  • Multiply the entire equation by the original denominator to clear the fractions and solve for the constants by comparing coefficients.
This method simplifies the integral, making it easier to evaluate using basic integration techniques.
Rational Function
Rational functions are expressions of the form \(\frac{P(x)}{Q(x)}\) where \(P(x)\) and \(Q(x)\) are polynomials. The integrals of these functions can be tricky to solve directly but understanding their properties makes them more manageable.

In the exercise presented:
  • \(P(x) = 4x - 2\) and \(Q(x) = x^2 - 9\). The degree of \(Q(x)\) is higher than that of \(P(x)\), confirming it as a proper rational function.
  • The process of partial fraction decomposition turns the challenging fraction into simpler parts.
This transformation breaks the integral into smaller, familiar forms which are solveable using standard techniques such as integration formulas. Understanding the structure of rational functions is essential to apply such techniques efficiently.
Integration Techniques
Integration techniques are strategies to find antiderivatives or evaluate integrals. In this scenario, after applying partial fraction decomposition, integration becomes straightforward:

We integrate each simple fraction separately. Consider our example:
  • The integral \(\int \frac{5/3}{x-3} \, dx\) simplifies to \(\frac{5}{3} \ln |x-3| + C_1\).
  • Similarly, \(\int \frac{7/3}{x+3} \, dx\) yields \(\frac{7}{3} \ln |x+3| + C_2\).
These are derived using the standard integral form \(\int \frac{k}{x-a} \, dx = k \ln |x-a| + C\).

Each time you identify a partial fraction, integrate using these simpler forms. By breaking the complex integral into these manageable components, integration becomes a series of straightforward steps.

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Most popular questions from this chapter

(a) Between 2005 and \(2015,\) ACME Widgets sold widgets at a continuous rate of \(R=R_{0} e^{0.125 t}\) widgets per year, where \(t\) is time in years since January 1 2005\. Suppose they were selling widgets at a rate of 1000 per year on January \(1,2005 .\) How many widgets did they sell between 2005 and \(2015 ?\) How many did they sell if the rate on January 1,2005 was 1,000,000 widgets per year? (b) In the first case ( 1000 widgets per year on January 1,2005)\(,\) how long did it take for half the widgets in the ten-year period to be sold? In the second case \((1.000 .000 \text { widgets per year on January } 1,2005)\) when had half the widgets in the ten-year period been sold? (c) In \(2015,\) ACME advertised that half the widgets it had sold in the previous ten years were still in use. Based on your answer to part (b), how long must a widget last in order to justify this claim?

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