We have determined that A = 36 and B = -6. Therefore, we have the following. A 6x + 1 30 (6x + 1)(x + 1) 36 6x + 1 x 36 = x 6 x + 1 6 x + 1 This allows us to rewrite the definite integral as follows. 1 36 6 [² [²(6x + 1 = x + 1)² dx dx = 30 6x² + 7x + 1 Finally, evaluating this integral gives the following result. (Remember to use absolute values where appropriate.) x + 1) αx² + dx = B x + 1 = 10] - 6 In(x + 1)
We have determined that A = 36 and B = -6. Therefore, we have the following. A 6x + 1 30 (6x + 1)(x + 1) 36 6x + 1 x 36 = x 6 x + 1 6 x + 1 This allows us to rewrite the definite integral as follows. 1 36 6 [² [²(6x + 1 = x + 1)² dx dx = 30 6x² + 7x + 1 Finally, evaluating this integral gives the following result. (Remember to use absolute values where appropriate.) x + 1) αx² + dx = B x + 1 = 10] - 6 In(x + 1)
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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![**Partial Fraction Decomposition and Integration**
To solve the integral involving the rational function \(\frac{30}{6x^2 + 7x + 1}\), we start by applying partial fraction decomposition. The general form for the decomposition of our expression is:
\[
\frac{30}{(6x + 1)(x + 1)} = \frac{A}{6x + 1} + \frac{B}{x + 1}
\]
We have determined from comparison and solving that \(A = 36\) and \(B = -6\). Therefore, we can rewrite the rational function as follows:
\[
\frac{30}{(6x + 1)(x + 1)} = \frac{36}{6x+1} - \frac{6}{x+1}
\]
This allows us to rewrite the definite integral as follows:
\[
\int_{0}^{1} \frac{30}{6x^2 + 7x + 1} \, dx = \int_{0}^{1} \left( \frac{36}{6x+1} - \frac{6}{x+1} \right) \, dx
\]
To integrate, we separate the integral into two simpler integrals:
\[
\int_{0}^{1} \left( \frac{36}{6x+1} - \frac{6}{x+1} \right) \, dx
\]
Finally, evaluating this integral yields the following expression (remember to use absolute values where appropriate):
\[
\int_{0}^{1} \left( \frac{36}{6x+1} - \frac{6}{x+1} \right) \, dx = \left[ 6 \ln |6x+1| - 6 \ln |x+1| \right]_0^1
\]
Evaluating at the bounds 0 and 1:
\[
= \left( 6 \ln |7| - 6 \ln |2| \right) - \left( 6 \ln |1| - 6 \ln |1| \right)
\]
Since \(\ln(1)=0\), this simplifies to:
\[
= 6 \ln |7| - 6 \ln |2|
\]
Therefore, the final evaluated integral is:
\[](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc4971f23-5570-4fed-8cd6-5e1a07d18a42%2F7fd8e47e-6e4f-40a9-a6ff-a65aa1cbe5be%2Fh1phuzc_processed.png&w=3840&q=75)
Transcribed Image Text:**Partial Fraction Decomposition and Integration**
To solve the integral involving the rational function \(\frac{30}{6x^2 + 7x + 1}\), we start by applying partial fraction decomposition. The general form for the decomposition of our expression is:
\[
\frac{30}{(6x + 1)(x + 1)} = \frac{A}{6x + 1} + \frac{B}{x + 1}
\]
We have determined from comparison and solving that \(A = 36\) and \(B = -6\). Therefore, we can rewrite the rational function as follows:
\[
\frac{30}{(6x + 1)(x + 1)} = \frac{36}{6x+1} - \frac{6}{x+1}
\]
This allows us to rewrite the definite integral as follows:
\[
\int_{0}^{1} \frac{30}{6x^2 + 7x + 1} \, dx = \int_{0}^{1} \left( \frac{36}{6x+1} - \frac{6}{x+1} \right) \, dx
\]
To integrate, we separate the integral into two simpler integrals:
\[
\int_{0}^{1} \left( \frac{36}{6x+1} - \frac{6}{x+1} \right) \, dx
\]
Finally, evaluating this integral yields the following expression (remember to use absolute values where appropriate):
\[
\int_{0}^{1} \left( \frac{36}{6x+1} - \frac{6}{x+1} \right) \, dx = \left[ 6 \ln |6x+1| - 6 \ln |x+1| \right]_0^1
\]
Evaluating at the bounds 0 and 1:
\[
= \left( 6 \ln |7| - 6 \ln |2| \right) - \left( 6 \ln |1| - 6 \ln |1| \right)
\]
Since \(\ln(1)=0\), this simplifies to:
\[
= 6 \ln |7| - 6 \ln |2|
\]
Therefore, the final evaluated integral is:
\[
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