Express 9x/(2x+1)(x²+1) in partial fraction

To express \(\frac{9x}{(2x+1)(x^2+1)}\) as a sum of partial fractions, we follow these steps:

### Step 1: Set up the partial fractions
We assume the partial fraction decomposition takes the form:

\[
\frac{9x}{(2x+1)(x^2+1)} = \frac{A}{2x+1} + \frac{Bx + C}{x^2+1}
\]

### Step 2: Multiply through by the common denominator
Multiply both sides by \((2x+1)(x^2+1)\) to eliminate the denominators:

\[
9x = A(x^2 + 1) + (Bx + C)(2x + 1)
\]

### Step 3: Expand and collect like terms
Expand the right-hand side:

\[
9x = A(x^2 + 1) + (Bx + C)(2x + 1)
\]

Expand each term:

\[
9x = A(x^2) + A(1) + Bx(2x) + Bx(1) + C(2x) + C(1)
\]

\[
9x = Ax^2 + A + 2Bx^2 + Bx + 2Cx + C
\]

Now, collect like terms:

\[
9x = (A + 2B)x^2 + (B + 2C)x + (A + C)
\]

### Step 4: Match coefficients
Now, match the coefficients of corresponding powers of \(x\) on both sides:

For \(x^2\):
\[
A + 2B = 0
\]

For \(x\):
\[
B + 2C = 9
\]

For the constant term:
\[
A + C = 0
\]

### Step 5: Solve the system of equations
We now solve this system of equations:

1. From \(A + 2B = 0\), we have \(A = -2B\).
2. Substitute \(A = -2B\) into \(A + C = 0\) to get \(-2B + C = 0\), so \(C = 2B\).
3. Substitute \(C = 2B\) into \(B + 2C = 9\) to get \(B + 4B = 9\), so \(5B = 9\), which gives \(B = \frac{9}{5}\).
4. Since \(C = 2B\), we get \(C = 2 \times \frac{9}{5} = \frac{18}{5}\).
5. Since \(A = -2B\), we get \(A = -2 \times \frac{9}{5} = -\frac{18}{5}\).

### Step 6: Write the final partial fraction decomposition
Now that we have \(A\), \(B\), and \(C\), substitute them back into the partial fraction form:

\[
\frac{9x}{(2x+1)(x^2+1)} = \frac{-\frac{18}{5}}{2x+1} + \frac{\frac{9}{5}x + \frac{18}{5}}{x^2+1}
\]

This can be simplified to:

\[
\frac{9x}{(2x+1)(x^2+1)} = -\frac{\frac{18}{5}}{2x+1} + \frac{\frac{9}{5}(x+2)}{x^2+1}
\]

This is the partial fraction decomposition of the given expression.

Let me know if you need any further details or have any questions.

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Here are five related questions:
1. How do you perform partial fraction decomposition when dealing with repeated factors?
2. Can partial fractions be applied to improper fractions? If so, how?
3. What are the applications of partial fractions in solving integrals?
4. How do you solve a system of equations with three or more variables?
5. How do partial fractions help in simplifying complex rational functions?

**Tip:** When solving partial fraction problems, always double-check your coefficients and expand the terms to ensure the decomposition is correct.

Learn how to perform partial fraction decomposition on the rational expression 9x/(2x+1)(x²+1). This detailed solution covers the steps involved in decomposing the expression into partial fractions, solving a system of equations to find the coefficients, and simplifying the result. Suitable for advanced high school students in Grades 11-12.

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