Ax + B Cx+D Then (22 + 3)2 8.5.57 (2² +3)2 x2 +3 2 + 5z = (Az + B)(x² + 3) + Cz + D = Ar + Bx² + (3.A + C)r +3B+ D. Then A = 1, B = 0, 3A+ C = 5, and 3B+ D = 0. So C= 2 and D=0. We have + 5x 2.r In a° + 3| - (x² + 3)² dr = (22 +3)2 dx = +C. r2 2+3 +3
Ax + B Cx+D Then (22 + 3)2 8.5.57 (2² +3)2 x2 +3 2 + 5z = (Az + B)(x² + 3) + Cz + D = Ar + Bx² + (3.A + C)r +3B+ D. Then A = 1, B = 0, 3A+ C = 5, and 3B+ D = 0. So C= 2 and D=0. We have + 5x 2.r In a° + 3| - (x² + 3)² dr = (22 +3)2 dx = +C. r2 2+3 +3
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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How do we solve for B and D in this problem? I was able to get A and C but couldn't figure out B And D
![The image displays a mathematical solution involving partial fraction decomposition and integration.
The problem is labeled as "8.5.57" and involves expressing the fraction \(\frac{x^3 + 5x}{(x^2+3)^2}\) as a sum of simpler fractions:
\[
\frac{x^3 + 5x}{(x^2+3)^2} = \frac{Ax+B}{x^2+3} + \frac{Cx+D}{(x^2+3)^2}
\]
To solve for \(A\), \(B\), \(C\), and \(D\), we equate:
\[
x^3 + 5x = (Ax + B)(x^2 + 3) + Cx + D = Ax^3 + Bx^2 + (3A + C)x + 3B + D
\]
By comparing coefficients, we find:
- \(A = 1\)
- \(B = 0\)
- \(3A + C = 5\)
- \(3B + D = 0\)
Thus, solving these gives:
- \(C = 2\)
- \(D = 0\)
The integral of the original function is then evaluated:
\[
\int \frac{x^3 + 5x}{(x^2+3)^2} \, dx = \int \left( \frac{x}{x^2+3} + \frac{2x}{(x^2+3)^2} \right) \, dx
\]
This results in:
\[
= \frac{1}{2} \ln |x^2+3| - \frac{1}{x^2+3} + C
\]
This step-by-step process explains how the integration is carried out.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F61b75115-d70f-4fe2-af93-2076876ad69a%2F159629ab-8a7e-445c-bf12-3eec7247244f%2F4ird72i_processed.jpeg&w=3840&q=75)
Transcribed Image Text:The image displays a mathematical solution involving partial fraction decomposition and integration.
The problem is labeled as "8.5.57" and involves expressing the fraction \(\frac{x^3 + 5x}{(x^2+3)^2}\) as a sum of simpler fractions:
\[
\frac{x^3 + 5x}{(x^2+3)^2} = \frac{Ax+B}{x^2+3} + \frac{Cx+D}{(x^2+3)^2}
\]
To solve for \(A\), \(B\), \(C\), and \(D\), we equate:
\[
x^3 + 5x = (Ax + B)(x^2 + 3) + Cx + D = Ax^3 + Bx^2 + (3A + C)x + 3B + D
\]
By comparing coefficients, we find:
- \(A = 1\)
- \(B = 0\)
- \(3A + C = 5\)
- \(3B + D = 0\)
Thus, solving these gives:
- \(C = 2\)
- \(D = 0\)
The integral of the original function is then evaluated:
\[
\int \frac{x^3 + 5x}{(x^2+3)^2} \, dx = \int \left( \frac{x}{x^2+3} + \frac{2x}{(x^2+3)^2} \right) \, dx
\]
This results in:
\[
= \frac{1}{2} \ln |x^2+3| - \frac{1}{x^2+3} + C
\]
This step-by-step process explains how the integration is carried out.
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