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Chapter 12: Problem 22

Use partial fractions to find the indefinite integral. $$ \int \frac{1}{4 x^{2}-9} d x $$

Short Answer

Expert verified
The indefinite integral of \( \frac{1}{4x^{2}-9} \) is \( \frac{1}{5} \ln |2x - 3| - \frac{1}{5} \ln |2x + 3| + C \)

Step by step solution

01

Factor the Denominator

Factoring the denominator yields: \( 4x^{2}-9 = 2x - 3)(2x + 3) \)
02

Set Up the Partial Fraction Decomposition

\( \frac{1}{4x^{2}-9} = \frac{A}{2x - 3} + \frac{B}{2x + 3} \). Our task is to find the parameters A and B.
03

Solve for Parameters A and B

Multiply both sides by \(4x^{2}-9=(2x - 3)(2x + 3)\) to clear the fraction. This yields: \(1 = A(2x + 3) + B(2x - 3)\). Solve this equation for A and B by choosing smart x-values where A or B cancel. If you choose \(x=\frac{3}{2}\), you get \(1=5A \Rightarrow A=\frac{1}{5}\). If you choose \(x=-\frac{3}{2}\), you get \(1=-5B \Rightarrow B=-\frac{1}{5}\). So \( \frac{1}{4x^{2}-9} = \frac{\frac{1}{5}}{2x - 3} - \frac{\frac{1}{5}}{2x + 3} \)
04

Integrate Each Term

The integral of a sum is the sum of the integrals, allowing each term to be integrated separately. The integral of each term is a natural logarithm, yielding the final answer: \( \int \frac{1}{4 x^{2}-9} dx = \frac{1}{5} \ln |2x - 3| - \frac{1}{5} \ln |2x + 3| + C \)

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Most popular questions from this chapter

Present Value A business is expected to yield a continuous flow of profit at the rate of \(\$ 500,000\) per year. If money will earn interest at the nominal rate of \(9 \%\) per year compounded continuously, what is the present value of the business (a) for 20 years and (b) forever?

Use a program similar to the Simpson's Rule program on page 906 with \(n=6\) to approximate the indicated normal probability. The standard normal probability density function is \(f(x)=(1 / \sqrt{2 \pi}) e^{-x^{2} / 2}\). If \(x\) is chosen at random from a population with this density, then the probability that \(x\) lies in the interval \([a, b]\) is \(P(a \leq x \leq b)=\int_{a}^{b} f(x) d x\). $$ P(0 \leq x \leq 1.5) $$

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