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
I have been doing this question and I am lost as there is still something else that need to do to get the final answer,
can you help me.
![**Evaluate the Integral**
Given the integral:
\[
\int \frac{dt}{t^2 + t - 30}
\]
**Answer:**
[Provide space here for students to enter their response]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fdd3494d7-16cb-4f96-a1a3-f69334a1e65b%2Fae3859b2-c673-4e43-a5c0-8876096ecc45%2Ffonfjk_processed.png&w=3840&q=75)
Transcribed Image Text:**Evaluate the Integral**
Given the integral:
\[
\int \frac{dt}{t^2 + t - 30}
\]
**Answer:**
[Provide space here for students to enter their response]
![**Question 8: Evaluate**
\[
\int \frac{dt}{t^2 + t - 30}
\]
Rewrite the integrand:
\[
\frac{1}{t^2 + t - 30}
\]
Factor the quadratic expression:
\[
t^2 + t - 30 = (t + 6)(t - 5)
\]
Set up partial fractions:
\[
\frac{1}{(t-5)(t+6)} = \frac{A}{t-5} + \frac{B}{t+6}
\]
Multiply through by the least common denominator (LCD):
\[
A(t+6) + B(t-5) = 1
\]
Substitute convenient values for \(t\):
For \(t = 5\):
\[
A(5+6) = 1 \Rightarrow 11A = 1 \Rightarrow A = \frac{1}{11}
\]
For \(t = -6\):
\[
B(-6-5) = 1 \Rightarrow -11B = 1 \Rightarrow B = -\frac{1}{11}
\]
Substitute back into the integral:
\[
\int \frac{1}{11(t-5)} dt - \int \frac{1}{11(t+6)} dt
\]
Integrate each term:
\[
\frac{1}{11} \ln |t-5| - \frac{1}{11} \ln |t+6| + C
\]
Use the properties of logarithms to combine:
\[
\frac{1}{11} \left( \ln |t-5| - \ln |t+6| \right) + C = \frac{1}{11} \ln \left| \frac{t-5}{t+6} \right| + C
\]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fdd3494d7-16cb-4f96-a1a3-f69334a1e65b%2Fae3859b2-c673-4e43-a5c0-8876096ecc45%2F4l8y3a_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Question 8: Evaluate**
\[
\int \frac{dt}{t^2 + t - 30}
\]
Rewrite the integrand:
\[
\frac{1}{t^2 + t - 30}
\]
Factor the quadratic expression:
\[
t^2 + t - 30 = (t + 6)(t - 5)
\]
Set up partial fractions:
\[
\frac{1}{(t-5)(t+6)} = \frac{A}{t-5} + \frac{B}{t+6}
\]
Multiply through by the least common denominator (LCD):
\[
A(t+6) + B(t-5) = 1
\]
Substitute convenient values for \(t\):
For \(t = 5\):
\[
A(5+6) = 1 \Rightarrow 11A = 1 \Rightarrow A = \frac{1}{11}
\]
For \(t = -6\):
\[
B(-6-5) = 1 \Rightarrow -11B = 1 \Rightarrow B = -\frac{1}{11}
\]
Substitute back into the integral:
\[
\int \frac{1}{11(t-5)} dt - \int \frac{1}{11(t+6)} dt
\]
Integrate each term:
\[
\frac{1}{11} \ln |t-5| - \frac{1}{11} \ln |t+6| + C
\]
Use the properties of logarithms to combine:
\[
\frac{1}{11} \left( \ln |t-5| - \ln |t+6| \right) + C = \frac{1}{11} \ln \left| \frac{t-5}{t+6} \right| + C
\]
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