To find the partial fraction decomposition of the given rational expression:

$\frac{X^2}{(2x+1)(x+3)^2}$

we need to express it as a sum of simpler fractions. The general form for the partial fraction decomposition is:

$\frac{X^2}{(2x+1)(x+3)^2} = \frac{A}{2x+1} + \frac{B}{x+3} + \frac{C}{(x+3)^2}$

Here, $A$, $B$, and $C$ are constants that we need to determine.

Steps to find the constants $A$, $B$, and $C$:

1. Multiply both sides by $(2x+1)(x+3)^2$ to clear the denominators:

   $X^2 = A(x+3)^2 + B(2x+1)(x+3) + C(2x+1)$

2. Expand the right-hand side:

   Expand each term on the right:

   $A(x+3)^2 = A(x^2 + 6x + 9)$

   $B(2x+1)(x+3) = B(2x^2 + 7x + 3)$

   $C(2x+1) = 2Cx + C$

   Therefore,

   $X^2 = A(x^2 + 6x + 9) + B(2x^2 + 7x + 3) + C(2x + 1)$

3. Combine like terms:

   $X^2 = (A + 2B)x^2 + (6A + 7B + 2C)x + (9A + 3B + C)$

4. Compare the coefficients on both sides of the equation:

   Compare the coefficients of $x^2$, $x$, and the constant term from both sides:

   $\text{For } x^2: \quad 1 = A + 2B$

   $\text{For } x: \quad 0 = 6A + 7B + 2C$

   $\text{For the constant term:} \quad 0 = 9A + 3B + C$

5. Solve the system of equations:

   Solve the system of linear equations to find the values of $A$, $B$, and $C$.

Let me calculate the values of $A$, $B$, and $C$ for you.It seems there was a small typo in the code. Let me correct that and solve the system of equations again.The issue lies in how the equations were written. I’ll correct it and solve the system.The solution to the system of equations is:

$A = \frac{1}{25}, \quad B = \frac{12}{25}, \quad C = -\frac{9}{5}$

So, the partial fraction decomposition of the given expression is:

$\frac{X^2}{(2x+1)(x+3)^2} = \frac{1/25}{2x+1} + \frac{12/25}{x+3} - \frac{9/5}{(x+3)^2}$

Would you like further details on this process, or do you have any questions?

Related Questions:

1. How can we verify the partial fraction decomposition?
2. What are partial fractions used for in calculus?
3. Can this method be applied to improper fractions?
4. How do we handle repeated irreducible factors in partial fractions?
5. What happens if the denominator has an irreducible quadratic factor?
6. How can partial fractions be used to integrate rational functions?
7. How does this method change if the numerator has a higher degree than the denominator?
8. What is the connection between partial fractions and Laplace transforms?

Tip: When solving partial fractions, always ensure that the degree of the numerator is less than the degree of the denominator before decomposing.