If `(x^2)/((x^(2)+1)(x^(2)+4))=(a)/(x^(2)+1)+(b)/(x^(2)+4)` then `a+b=`

Options

A. `0`
B. `1`
C. `2`
D. `3`

Answer

To solve the equation

\[ \frac{x^2}{(x^2 + 1)(x^2 + 4)} = \frac{a}{x^2 + 1} + \frac{b}{x^2 + 4} \]

we will use the method of partial fractions. Here are the steps:

### Step 1: Rewrite the equation
Let \( t = x^2 \). Then, we can rewrite the equation as:

\[ \frac{t}{(t + 1)(t + 4)} = \frac{a}{t + 1} + \frac{b}{t + 4} \]

### Step 2: Find a common denominator
The common denominator on the right-hand side is \( (t + 1)(t + 4) \). Thus, we can write:

\[ \frac{a(t + 4) + b(t + 1)}{(t + 1)(t + 4)} = \frac{t}{(t + 1)(t + 4)} \]

### Step 3: Equate the numerators
Since the denominators are equal, we can equate the numerators:

\[ t = a(t + 4) + b(t + 1) \]

### Step 4: Expand the right-hand side
Expanding the right-hand side gives:

\[ t = at + 4a + bt + b \]

Combining like terms, we have:

\[ t = (a + b)t + (4a + b) \]

### Step 5: Set up equations
Now, we can compare coefficients from both sides. From the equation:

\[ t = (a + b)t + (4a + b) \]

we can derive two equations:

1. Coefficient of \( t \): \( 1 = a + b \)
2. Constant term: \( 0 = 4a + b \)

### Step 6: Solve the system of equations
From the first equation, we have:

\[ b = 1 - a \]

Substituting \( b \) in the second equation:

\[ 0 = 4a + (1 - a) \]
\[ 0 = 4a + 1 - a \]
\[ 0 = 3a + 1 \]
\[ 3a = -1 \quad \Rightarrow \quad a = -\frac{1}{3} \]

Now substituting \( a \) back to find \( b \):

\[ b = 1 - \left(-\frac{1}{3}\right) = 1 + \frac{1}{3} = \frac{4}{3} \]

### Step 7: Calculate \( a + b \)
Now, we can find \( a + b \):

\[ a + b = -\frac{1}{3} + \frac{4}{3} = \frac{3}{3} = 1 \]

Thus, the final answer is:

\[ \boxed{1} \]