If `int(2x^(2)+5)/(x^(2)+a)dx=f(x)`, where f(x) is a polynomial or ratio of polynomials, then the number of possible value(s) of a is equal to

Options

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

Answer

To solve the problem, we need to analyze the integral given:

\[ \int \frac{2x^2 + 5}{x^2 + a} \, dx = f(x) \]

where \( f(x) \) is a polynomial or a ratio of polynomials.

### Step 1: Rewrite the Integral
We can rewrite the integrand by separating the terms:

\[ \frac{2x^2 + 5}{x^2 + a} = \frac{2x^2 + 2a - 2a + 5}{x^2 + a} = \frac{2(x^2 + a) + (5 - 2a)}{x^2 + a} \]

### Step 2: Split the Integral
Now, we can split the integral into two parts:

\[ \int \frac{2(x^2 + a)}{x^2 + a} \, dx + \int \frac{5 - 2a}{x^2 + a} \, dx \]

This simplifies to:

\[ 2 \int dx + (5 - 2a) \int \frac{1}{x^2 + a} \, dx \]

### Step 3: Evaluate the First Integral
The first integral is straightforward:

\[ 2 \int dx = 2x \]

### Step 4: Evaluate the Second Integral
The second integral, \( \int \frac{1}{x^2 + a} \, dx \), depends on the value of \( a \). If \( a > 0 \), it can be expressed as:

\[ \int \frac{1}{x^2 + a} \, dx = \frac{1}{\sqrt{a}} \tan^{-1}\left(\frac{x}{\sqrt{a}}\right) + C \]

### Step 5: Analyze the Result
Thus, the integral becomes:

\[ 2x + (5 - 2a) \left( \frac{1}{\sqrt{a}} \tan^{-1}\left(\frac{x}{\sqrt{a}}\right) \right) + C \]

For \( f(x) \) to be a polynomial or a ratio of polynomials, the term \( (5 - 2a) \) must equal zero, because the term involving \( \tan^{-1} \) would otherwise introduce a non-polynomial component.

### Step 6: Solve for \( a \)
Setting \( 5 - 2a = 0 \):

\[ 5 = 2a \implies a = \frac{5}{2} \]

### Conclusion
The only possible value of \( a \) that allows \( f(x) \) to remain a polynomial or a ratio of polynomials is:

\[ a = \frac{5}{2} \]

Thus, the number of possible values of \( a \) is **1**.