(x - 1)(x² +64) (x - 1)(x² +64) Setting the numerators equal we can determine the following. A+B=0✔ 0 C-B- 0 0 64AC = 65✓ 65 Step 4 The equation 64A - C= 65 is in terms of A and C. To simplify the process of solving, we need a second equation in terms of A and C. Since CB= 0, we know that C= B. Therefore, we can rewrite the equation A + B = 0 using substitution so that it is in terms of A and C as follows. A+C =0 We have a set of equations. A+C =0 64A - C = 65 Solving for A and C gives the following result. A = 11 C = -1✔ Further, since C= B we can also find B. B=-1✔ -1 Step 5 We can now apply the values A 1, B-1, and C -1. A x-1 + Bx + C x² + 64 - Therefore, we have the following. 65 (x - 1)(x² +64) X-X1 x-1 dx = + - dx = ‹ - / ( √ + + + + - -X- -1 x + (-1✓ x+64 X + C x² + 64 X - √(x = 1= x ² + [64] ✓ 64 dx -1 Step 6 Using the substitution u = x - 1, we get the following. (Remember to use absolute values where appropriate.) 11 x2+64 dx
(x - 1)(x² +64) (x - 1)(x² +64) Setting the numerators equal we can determine the following. A+B=0✔ 0 C-B- 0 0 64AC = 65✓ 65 Step 4 The equation 64A - C= 65 is in terms of A and C. To simplify the process of solving, we need a second equation in terms of A and C. Since CB= 0, we know that C= B. Therefore, we can rewrite the equation A + B = 0 using substitution so that it is in terms of A and C as follows. A+C =0 We have a set of equations. A+C =0 64A - C = 65 Solving for A and C gives the following result. A = 11 C = -1✔ Further, since C= B we can also find B. B=-1✔ -1 Step 5 We can now apply the values A 1, B-1, and C -1. A x-1 + Bx + C x² + 64 - Therefore, we have the following. 65 (x - 1)(x² +64) X-X1 x-1 dx = + - dx = ‹ - / ( √ + + + + - -X- -1 x + (-1✓ x+64 X + C x² + 64 X - √(x = 1= x ² + [64] ✓ 64 dx -1 Step 6 Using the substitution u = x - 1, we get the following. (Remember to use absolute values where appropriate.) 11 x2+64 dx
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
please help with step 6
![**Step 3**
We can now state the following:
\[ \frac{65}{(x - 1)(x^2 + 64)} = \frac{(A + B)x^2 + (C - B)x + (64A - C)}{(x - 1)(x^2 + 64)} \]
Setting the numerators equal we can determine the following.
\[
\begin{align*}
A + B &= 0 \quad \textcolor{#008000}{\checkmark \boxed{0}} \\
C - B &= 0 \quad \textcolor{#008000}{\checkmark \boxed{0}} \\
64A - C &= 65 \quad \textcolor{#008000}{\checkmark \boxed{65}}
\end{align*}
\]
**Step 4**
The equation \(64A - C = 65\) is in terms of \(A\) and \(C\). To simplify the process of solving, we need a second equation in terms of \(A\) and \(C\).
Since \(C - B = 0\), we know that \(C = B\). Therefore, we can rewrite the equation \(A + B = 0\) using substitution so that it is in terms of \(A\) and \(C\) as follows.
\[ A + C = 0 \]
We have a set of equations.
\[
\begin{align*}
A + C &= 0 \\
64A - C &= 65
\end{align*}
\]
Solving for \(A\) and \(C\) gives the following result.
\[
\begin{align*}
A &= \frac{1}{1} \quad \textcolor{#008000}{\checkmark \boxed{1}} \\
C &= -1 \quad \textcolor{#008000}{\checkmark \boxed{-1}}
\end{align*}
\]
Further, since \(C = B\) we can also find \(B\).
\[ B = -1 \quad \textcolor{#008000}{\checkmark \boxed{-1}} \]
**Step 5**
We can now apply the values \(A = 1\), \(B = -1\), and \(C = -1\).
\[
\frac{A}{x](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc4971f23-5570-4fed-8cd6-5e1a07d18a42%2F54d571e7-fd3b-4cde-b66d-41ec7afa9a5e%2F8essttj_processed.png&w=3840&q=75)
Transcribed Image Text:**Step 3**
We can now state the following:
\[ \frac{65}{(x - 1)(x^2 + 64)} = \frac{(A + B)x^2 + (C - B)x + (64A - C)}{(x - 1)(x^2 + 64)} \]
Setting the numerators equal we can determine the following.
\[
\begin{align*}
A + B &= 0 \quad \textcolor{#008000}{\checkmark \boxed{0}} \\
C - B &= 0 \quad \textcolor{#008000}{\checkmark \boxed{0}} \\
64A - C &= 65 \quad \textcolor{#008000}{\checkmark \boxed{65}}
\end{align*}
\]
**Step 4**
The equation \(64A - C = 65\) is in terms of \(A\) and \(C\). To simplify the process of solving, we need a second equation in terms of \(A\) and \(C\).
Since \(C - B = 0\), we know that \(C = B\). Therefore, we can rewrite the equation \(A + B = 0\) using substitution so that it is in terms of \(A\) and \(C\) as follows.
\[ A + C = 0 \]
We have a set of equations.
\[
\begin{align*}
A + C &= 0 \\
64A - C &= 65
\end{align*}
\]
Solving for \(A\) and \(C\) gives the following result.
\[
\begin{align*}
A &= \frac{1}{1} \quad \textcolor{#008000}{\checkmark \boxed{1}} \\
C &= -1 \quad \textcolor{#008000}{\checkmark \boxed{-1}}
\end{align*}
\]
Further, since \(C = B\) we can also find \(B\).
\[ B = -1 \quad \textcolor{#008000}{\checkmark \boxed{-1}} \]
**Step 5**
We can now apply the values \(A = 1\), \(B = -1\), and \(C = -1\).
\[
\frac{A}{x
![**Evaluate the integral.**
\[
\int \frac{65}{(x-1)(x^2 + 64)} \, dx
\]
---
**Step 1**
The given function \(\int \frac{65}{(x-1)(x^2 + 64)} \, dx\) is a rational function of the form \(f(x) = \frac{P(x)}{Q(x)}\) where \(P(x) = 65\) and \(Q(x) = (x-1)(x^2 + 64)\). Since the degree of the numerator \(P(x)\) is less than the degree of the denominator \(Q(x)\), we do not need to divide. We also note that we cannot factor the denominator more than it is already factored. We further note that \(Q(x)\) contains irreducible quadratic factors, none of which are repeated. This allows us to find a partial fraction decomposition of the integrand as follows.
\[
\frac{65}{(x-1)(x^2 + 64)} = \frac{A}{x-1} + \frac{Bx + C}{x^2 + 64}
\]
\[
\frac{65}{(x-1)(x^2 + 64)} = \frac{A}{x-1} + \frac{Bx + C}{x^2 + 64}
\]
---
**Step 2**
Combining the fractions, we get:
\[
\frac{A}{x-1} + \frac{Bx + C}{x^2 + 64} = \frac{(A + Bx)(x^2 + 64) + (C - Bx)}{(x-1)(x^2 + 64)}
\]
\[
= \frac{(A + B)x^2 + (C - B)x + (64A - C)}{(x-1)(x^2 + 64)}
\]
---
**Step 3**
We can now state the following.
\[
\frac{65}{(x-1)(x^2 + 64)} = \frac{(A+B)x^2 + (C - B)x + (64A - C)}{(x-1)(x^2 + 64)}
\]
Setting the numerators equal, we can determine the following equations](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc4971f23-5570-4fed-8cd6-5e1a07d18a42%2F54d571e7-fd3b-4cde-b66d-41ec7afa9a5e%2Fxuiif3r_processed.png&w=3840&q=75)
Transcribed Image Text:**Evaluate the integral.**
\[
\int \frac{65}{(x-1)(x^2 + 64)} \, dx
\]
---
**Step 1**
The given function \(\int \frac{65}{(x-1)(x^2 + 64)} \, dx\) is a rational function of the form \(f(x) = \frac{P(x)}{Q(x)}\) where \(P(x) = 65\) and \(Q(x) = (x-1)(x^2 + 64)\). Since the degree of the numerator \(P(x)\) is less than the degree of the denominator \(Q(x)\), we do not need to divide. We also note that we cannot factor the denominator more than it is already factored. We further note that \(Q(x)\) contains irreducible quadratic factors, none of which are repeated. This allows us to find a partial fraction decomposition of the integrand as follows.
\[
\frac{65}{(x-1)(x^2 + 64)} = \frac{A}{x-1} + \frac{Bx + C}{x^2 + 64}
\]
\[
\frac{65}{(x-1)(x^2 + 64)} = \frac{A}{x-1} + \frac{Bx + C}{x^2 + 64}
\]
---
**Step 2**
Combining the fractions, we get:
\[
\frac{A}{x-1} + \frac{Bx + C}{x^2 + 64} = \frac{(A + Bx)(x^2 + 64) + (C - Bx)}{(x-1)(x^2 + 64)}
\]
\[
= \frac{(A + B)x^2 + (C - B)x + (64A - C)}{(x-1)(x^2 + 64)}
\]
---
**Step 3**
We can now state the following.
\[
\frac{65}{(x-1)(x^2 + 64)} = \frac{(A+B)x^2 + (C - B)x + (64A - C)}{(x-1)(x^2 + 64)}
\]
Setting the numerators equal, we can determine the following equations
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