Chapter 7: Problem 71
Evaluate each integral. $$ \int \frac{1}{x^{2}-9} d x $$
Short Answer
Expert verified
\(\int \frac{1}{x^2 - 9} \, dx = \frac{1}{6} \ln \left| \frac{x-3}{x+3} \right| + C \).
Step by step solution
01
Recognize the Denominator as a Difference of Squares
The denominator in the integrand, \(x^2 - 9\), is a difference of squares, which can be factored as \((x - 3)(x + 3)\). This indicates that partial fraction decomposition could simplify the integration process.
02
Set Up Partial Fraction Decomposition
Decompose the integrand \(\frac{1}{x^2 - 9}\) into partial fractions. Assume \(\frac{1}{x^2 - 9} = \frac{A}{x-3} + \frac{B}{x+3}\), where \(A\) and \(B\) are constants to be determined.
03
Solve for the Constants in Partial Fractions
Multiply through by \((x - 3)(x + 3)\) and set equal to the numerator: \[ 1 = A(x + 3) + B(x - 3) \]Expand and group like terms:\[ 1 = (A + B)x + (3A - 3B) \]This results in the system of equations:- \(A + B = 0\)- \(3A - 3B = 1\)Solve this system to find the values of \(A\) and \(B\):- From \(A + B = 0\), we have \(B = -A\).- Substitute \(B = -A\) into the second equation: \[3A - 3(-A) = 1\] \[6A = 1\] \[A = \frac{1}{6}\] Hence, \(B = -\frac{1}{6}\).
04
Write the Integral in Terms of Partial Fractions
Substitute \(A\) and \(B\) back into the partial fraction decomposition:\[ \int \frac{1}{x^2 - 9} \, dx = \int \left( \frac{1/6}{x-3} - \frac{1/6}{x+3} \right) \, dx \]
05
Integrate Each Term Individually
Recognize that the two integrals are simple logarithmic forms:\[ \int \frac{1/6}{x-3} \, dx = \frac{1}{6} \ln |x-3| + C_1 \]\[ \int \frac{-1/6}{x+3} \, dx = -\frac{1}{6} \ln |x+3| + C_2 \]Combine these results to find:\[ \frac{1}{6} \ln |x-3| - \frac{1}{6} \ln |x+3| + C \]
06
Simplify the Result Using Logarithm Properties
Combine the logarithmic terms using properties of logarithms:\[ \frac{1}{6} (\ln |x-3| - \ln |x+3|) = \frac{1}{6} \ln \left| \frac{x-3}{x+3} \right| + C \]
07
Present the Final Answer
Thus, the final antiderivative of the original integral is:\[ \int \frac{1}{x^2 - 9} \, dx = \frac{1}{6} \ln \left| \frac{x-3}{x+3} \right| + C \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Integral Calculus
Integral calculus is a branch of mathematics that deals with finding the total value offset by a function, often over a certain range. One primary goal is calculating the area under a curve on a graph of a function. When we integrate a function, we're effectively summing an infinite number of infinitesimally small quantities to find a whole.
In this exercise, we looked at the integral \( \int \frac{1}{x^{2}-9} \, dx \), which involves finding the antiderivative of the given function. The goal was to transform this integral into a more manageable form using algebraic techniques such as partial fraction decomposition.
In this exercise, we looked at the integral \( \int \frac{1}{x^{2}-9} \, dx \), which involves finding the antiderivative of the given function. The goal was to transform this integral into a more manageable form using algebraic techniques such as partial fraction decomposition.
- Antiderivatives: These are functions that reverse the process of differentiation, providing the original function that when differentiated gives the integrand.
- Indefinite Integrals: Since there are infinitely many antiderivatives, the indefinite integral includes a constant of integration \(C\).
Difference of Squares
The concept of difference of squares is a useful algebraic identity often utilized in various math problems. It is especially pertinent when dealing with expressions like \(x^2 - 9\), which appear in our given integral.
The difference of squares follows the equation: \(a^2 - b^2 = (a-b)(a+b)\). For our integrand \(x^2 - 9\), we recognize it as a difference of squares where \(a = x\) and \(b = 3\). Thus, it can be factored into:
Using this identity can make complex integration problems much simpler, and is a staple technique in algebra and calculus for simplifying and solving equations that appear complicated at first glance.
The difference of squares follows the equation: \(a^2 - b^2 = (a-b)(a+b)\). For our integrand \(x^2 - 9\), we recognize it as a difference of squares where \(a = x\) and \(b = 3\). Thus, it can be factored into:
- \(x^2 - 3^2 = (x-3)(x+3)\)
Using this identity can make complex integration problems much simpler, and is a staple technique in algebra and calculus for simplifying and solving equations that appear complicated at first glance.
Logarithmic Integration
Logarithmic integration is a technique utilized when integrating functions involving \frac{1}{x} type expressions. When you encounter integrals such as \int \frac{1}{x} \, dx, the result is the natural logarithm of the absolute value of \(x\), represented as \(\ln |x| + C\).
In our exercise, after applying partial fraction decomposition, we ended up with simpler integrals of the form \(\int \frac{1/6}{x-3} \, dx\) and \(\int \frac{-1/6}{x+3} \, dx\). These integrals identified as logarithmic because they follow the aforementioned form.
In our exercise, after applying partial fraction decomposition, we ended up with simpler integrals of the form \(\int \frac{1/6}{x-3} \, dx\) and \(\int \frac{-1/6}{x+3} \, dx\). These integrals identified as logarithmic because they follow the aforementioned form.
- Application: Each term like \(\int \frac{1/6}{x-3} \, dx\) results in an expression \(\frac{1}{6} \ln |x - 3| + C\).
- Combining Logarithms: Using properties of logarithms like \(\ln a - \ln b = \ln \frac{a}{b}\), the expressions can be combined into one concise logarithmic expression.