If `(2x^(2)+5)/((x+1)^(2)(x-3))=(A)/(x+1)+(B)/((x+1)^(2))+(C)/(x-3)` then `A=`

Options

A. `- (15)/(4)`
B. `15/4`
C. `9/16`
D. `- 9/16`

Answer

To solve the equation

\[ \frac{2x^2 + 5}{(x+1)^2(x-3)} = \frac{A}{x+1} + \frac{B}{(x+1)^2} + \frac{C}{x-3} \]

we will follow these steps:

### Step 1: Set up the equation
We start with the left-hand side as it is and rewrite the right-hand side with a common denominator:

\[ \frac{A}{x+1} + \frac{B}{(x+1)^2} + \frac{C}{x-3} = \frac{A(x+1)(x-3) + B(x-3) + C(x+1)^2}{(x+1)^2(x-3)} \]

### Step 2: Cancel the denominators
Since the denominators are the same, we can equate the numerators:

\[ 2x^2 + 5 = A(x+1)(x-3) + B(x-3) + C(x+1)^2 \]

### Step 3: Expand the right-hand side
Now we will expand the right-hand side:

1. Expand \(A(x+1)(x-3)\):
\[ A(x^2 - 3x + x - 3) = A(x^2 - 2x - 3) \]

2. Expand \(B(x-3)\):
\[ Bx - 3B \]

3. Expand \(C(x+1)^2\):
\[ C(x^2 + 2x + 1) \]

Combining these, we have:

\[ 2x^2 + 5 = Ax^2 - 2Ax - 3A + Bx - 3B + Cx^2 + 2Cx + C \]

### Step 4: Combine like terms
Now, we will combine the terms on the right-hand side:

\[ ( A + C )x^2 + (-2A + B + 2C)x + (-3A - 3B + C) \]

### Step 5: Equate coefficients
Now we can equate coefficients from both sides:

1. Coefficient of \(x^2\):
\[ A + C = 2 \quad \text{(1)} \]

2. Coefficient of \(x\):
\[ -2A + B + 2C = 0 \quad \text{(2)} \]

3. Constant term:
\[ -3A - 3B + C = 5 \quad \text{(3)} \]

### Step 6: Solve for \(C\)
To find \(C\), we can substitute \(x = 3\) into the original equation:

\[ 2(3)^2 + 5 = A(3+1)(3-3) + B(3-3) + C(3+1)^2 \]

This simplifies to:

\[ 18 + 5 = C(4^2) \implies 23 = 16C \implies C = \frac{23}{16} \]

### Step 7: Substitute \(C\) back into equation (1)
Substituting \(C\) into equation (1):

\[ A + \frac{23}{16} = 2 \implies A = 2 - \frac{23}{16} \]

Finding a common denominator:

\[ A = \frac{32}{16} - \frac{23}{16} = \frac{9}{16} \]

### Final Answer
Thus, the value of \(A\) is

\[ \boxed{\frac{9}{16}} \]

---