Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Question
Use Trig Substitution.
*Note, don't forget that, x2 - a2 means that x=asectheta.
![### Integral of Rational Functions Involving Quadratic Terms
The image depicts the integral:
\[ \int \frac{1}{x^2 - a^2} \, dx \]
To solve this integral, we can use partial fraction decomposition and recognize that the integrand represents a function that can be broken down into simpler rational functions. Here’s the step-by-step process:
#### Step 1: Factor the Denominator
The integrand \(\frac{1}{x^2 - a^2}\) can be factored using the difference of squares:
\[ x^2 - a^2 = (x - a)(x + a) \]
So, we rewrite the integral as:
\[ \int \frac{1}{(x - a)(x + a)} \, dx \]
#### Step 2: Partial Fraction Decomposition
We express \(\frac{1}{(x - a)(x + a)}\) as a sum of partial fractions:
\[ \frac{1}{(x - a)(x + a)} = \frac{A}{x - a} + \frac{B}{x + a} \]
To determine \(A\) and \(B\), we will solve the equation:
\[ 1 = A(x + a) + B(x - a) \]
Setting \(x = a\) to solve for \(B\):
\[ 1 = A(a + a) + B(a - a) \]
\[ 1 = 2aA \]
\[ A = \frac{1}{2a} \]
Setting \(x = -a\) to solve for \(A\):
\[ 1 = A(-a + a) + B(-a - a) \]
\[ 1 = -2aB \]
\[ B = -\frac{1}{2a} \]
So, we have:
\[ \frac{1}{(x - a)(x + a)} = \frac{1}{2a} \cdot \frac{1}{x - a} - \frac{1}{2a} \cdot \frac{1}{x + a} \]
#### Step 3: Rewrite the Integral Using Partial Fractions
Now we can rewrite the integral:
\[ \int \frac{1}{(x - a)(x + a)} \, dx = \int \left( \frac{1}{2a](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff6a53db3-c060-4093-8251-3e3936495142%2F87d239a5-ff57-48fe-87eb-24ebe0d42b74%2Fi361i97_processed.png&w=3840&q=75)
Transcribed Image Text:### Integral of Rational Functions Involving Quadratic Terms
The image depicts the integral:
\[ \int \frac{1}{x^2 - a^2} \, dx \]
To solve this integral, we can use partial fraction decomposition and recognize that the integrand represents a function that can be broken down into simpler rational functions. Here’s the step-by-step process:
#### Step 1: Factor the Denominator
The integrand \(\frac{1}{x^2 - a^2}\) can be factored using the difference of squares:
\[ x^2 - a^2 = (x - a)(x + a) \]
So, we rewrite the integral as:
\[ \int \frac{1}{(x - a)(x + a)} \, dx \]
#### Step 2: Partial Fraction Decomposition
We express \(\frac{1}{(x - a)(x + a)}\) as a sum of partial fractions:
\[ \frac{1}{(x - a)(x + a)} = \frac{A}{x - a} + \frac{B}{x + a} \]
To determine \(A\) and \(B\), we will solve the equation:
\[ 1 = A(x + a) + B(x - a) \]
Setting \(x = a\) to solve for \(B\):
\[ 1 = A(a + a) + B(a - a) \]
\[ 1 = 2aA \]
\[ A = \frac{1}{2a} \]
Setting \(x = -a\) to solve for \(A\):
\[ 1 = A(-a + a) + B(-a - a) \]
\[ 1 = -2aB \]
\[ B = -\frac{1}{2a} \]
So, we have:
\[ \frac{1}{(x - a)(x + a)} = \frac{1}{2a} \cdot \frac{1}{x - a} - \frac{1}{2a} \cdot \frac{1}{x + a} \]
#### Step 3: Rewrite the Integral Using Partial Fractions
Now we can rewrite the integral:
\[ \int \frac{1}{(x - a)(x + a)} \, dx = \int \left( \frac{1}{2a
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