To solve the problem, we need to evaluate the integral
\[
\int \frac{2x^2 + 5}{x^2 + a} \, dx
\]
and determine the number of possible values of \(a\) such that the result is a polynomial or a ratio of polynomials.
### Step 1: Rewrite the integrand
We start by rewriting the integrand. We can express \(2x^2 + 5\) as follows:
\[
2x^2 + 5 = 2(x^2 + a) + (5 - 2a)
\]
This allows us to separate the integral into two parts:
\[
\int \frac{2x^2 + 5}{x^2 + a} \, dx = \int \frac{2(x^2 + a) + (5 - 2a)}{x^2 + a} \, dx
\]
### Step 2: Simplify the integral
Now we can split the integral:
\[
= \int \frac{2(x^2 + a)}{x^2 + a} \, dx + \int \frac{5 - 2a}{x^2 + a} \, dx
\]
The first integral simplifies to:
\[
\int 2 \, dx = 2x
\]
The second integral is:
\[
(5 - 2a) \int \frac{1}{x^2 + a} \, dx
\]
### Step 3: Evaluate the second integral
The integral \(\int \frac{1}{x^2 + a} \, dx\) can be evaluated as follows:
- If \(a > 0\), it results in a term involving \(\tan^{-1}\).
- If \(a = 0\), it leads to a logarithmic function.
- If \(a < 0\), it results in a different form.
### Step 4: Determine conditions for \(f(x)\) to be a polynomial
For \(f(x)\) to be a polynomial, the term \((5 - 2a)\) must equal zero, since the integral \(\int \frac{1}{x^2 + a} \, dx\) will not yield a polynomial if \(a > 0\) or \(a < 0\).
Setting \(5 - 2a = 0\):
\[
5 - 2a = 0 \implies 2a = 5 \implies a = \frac{5}{2}
\]
### Step 5: Check for other possible values of \(a\)
Additionally, if \(a = 0\), the integral simplifies to:
\[
\int \frac{2x^2 + 5}{x^2} \, dx = \int 2 + \frac{5}{x^2} \, dx
\]
This gives us a polynomial plus a term that is not a polynomial.
### Conclusion
Thus, the possible values of \(a\) that allow \(f(x)\) to be a polynomial or a ratio of polynomials are:
1. \(a = 0\)
2. \(a = \frac{5}{2}\)
Therefore, the number of possible values of \(a\) is:
\[
\boxed{2}
\]
