6 -dx Jx? - 5x+6 What integration technique would you use on Find the integral. Show every step.
6 -dx Jx? - 5x+6 What integration technique would you use on Find the integral. Show every step.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Question
![**Integration Technique Explanation:**
**Problem:**
Evaluate the integral:
\[
\int \frac{6}{x^2 - 5x + 6} \, dx
\]
**Solution:**
1. **Factor the Denominator:**
The denominator \(x^2 - 5x + 6\) can be factored as:
\[
x^2 - 5x + 6 = (x - 2)(x - 3)
\]
2. **Use Partial Fraction Decomposition:**
Express the fraction \(\frac{6}{(x - 2)(x - 3)}\) as a sum of partial fractions:
\[
\frac{6}{(x - 2)(x - 3)} = \frac{A}{x - 2} + \frac{B}{x - 3}
\]
Multiply through by the denominator \((x - 2)(x - 3)\) to clear the fractions:
\[
6 = A(x - 3) + B(x - 2)
\]
Expand and combine like terms:
\[
6 = Ax - 3A + Bx - 2B
\]
\[
6 = (A + B)x - (3A + 2B)
\]
By comparing coefficients, we have:
\[
A + B = 0
\]
\[
-3A - 2B = 6
\]
3. **Solve the System of Equations:**
From \(A + B = 0\), we get \(B = -A\).
Substitute \(B = -A\) into the second equation:
\[
-3A - 2(-A) = 6
\]
\[
-3A + 2A = 6
\]
\[
-A = 6 \implies A = -6
\]
Therefore, \(B = 6\).
4. **Integrate Each Term:**
Now rewrite the integral using the values of \(A\) and \(B\):
\[
\int \left(\frac{-6}{x - 2} + \frac{6}{x - 3}\](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff4586ed3-251d-413f-881b-72e0763cbb19%2Fa91c0016-7260-438a-b0d5-e98d47a1476d%2Fzf05pcs_processed.png&w=3840&q=75)
Transcribed Image Text:**Integration Technique Explanation:**
**Problem:**
Evaluate the integral:
\[
\int \frac{6}{x^2 - 5x + 6} \, dx
\]
**Solution:**
1. **Factor the Denominator:**
The denominator \(x^2 - 5x + 6\) can be factored as:
\[
x^2 - 5x + 6 = (x - 2)(x - 3)
\]
2. **Use Partial Fraction Decomposition:**
Express the fraction \(\frac{6}{(x - 2)(x - 3)}\) as a sum of partial fractions:
\[
\frac{6}{(x - 2)(x - 3)} = \frac{A}{x - 2} + \frac{B}{x - 3}
\]
Multiply through by the denominator \((x - 2)(x - 3)\) to clear the fractions:
\[
6 = A(x - 3) + B(x - 2)
\]
Expand and combine like terms:
\[
6 = Ax - 3A + Bx - 2B
\]
\[
6 = (A + B)x - (3A + 2B)
\]
By comparing coefficients, we have:
\[
A + B = 0
\]
\[
-3A - 2B = 6
\]
3. **Solve the System of Equations:**
From \(A + B = 0\), we get \(B = -A\).
Substitute \(B = -A\) into the second equation:
\[
-3A - 2(-A) = 6
\]
\[
-3A + 2A = 6
\]
\[
-A = 6 \implies A = -6
\]
Therefore, \(B = 6\).
4. **Integrate Each Term:**
Now rewrite the integral using the values of \(A\) and \(B\):
\[
\int \left(\frac{-6}{x - 2} + \frac{6}{x - 3}\
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