We are given the identity:

\( \frac{3}{(x+2)(x+1)} = \frac{a}{2x+1} + \frac{b}{x^2+1} \)

This is a partial fraction decomposition problem, but the given denominators are not factors of the denominator on the left-hand side. We can't directly apply the standard partial fraction method. Instead, let's analyze the equation. Notice that the denominators on the right are not factors of \((x+2)(x+1)\). This implies that a simple partial fraction decomposition is not possible in this form.

Let's consider a different approach. The given identity must hold true for all values of \(x\). Let's choose a convenient value of \(x\) to simplify the equation. Let's choose \(x=i\), where \(i\) is the imaginary unit (\(i^2 = -1\)). Substituting \(x = i\) into the equation, we get:

\( \frac{3}{(i+2)(i+1)} = \frac{a}{2i+1} + \frac{b}{i^2+1} \)

Since \(i^2 = -1\), the equation simplifies to:

\( \frac{3}{(i+2)(i+1)} = \frac{a}{2i+1} + \frac{b}{0} \)

For the equation to be defined, the term \( \frac{b}{0} \) must be eliminated which is possible only if \( b=0 \) . However, this contradicts the options given.

Let's try another approach. Multiplying both sides by \((x+2)(x+1)\) we have:

\( 3 = \frac{a(x+2)(x+1)}{2x+1} + \frac{b(x+2)(x+1)}{x^2+1} \)

Let's choose \(x=-1\). This will eliminate the first term on the right-hand side:

\(3 = 0 + \frac{b(-1+2)(-1+1)}{(-1)^2 + 1} = 0 \)

This is a contradiction. Let's choose \(x=-2\):

\(3 = 0 + \frac{b(-2+2)(-2+1)}{(-2)^2+1} = 0 \)

We have a contradiction. Let's assume that the question is wrongly stated and that the correct expression is

\( \frac{3}{(x+2)(x-1)} = \frac{a}{x+2} + \frac{b}{x-1} \)

Then using the standard partial fraction technique, we get a=1, b=-1. If the given identity holds true, there must be a misprint in the question.

Given the options, and assuming there's an error in the question, -1 seems to be the most plausible answer.