To solve the integral
\[
\int \frac{2x^2 + 3}{(x^2 - 1)(x^2 + 4)} \, dx,
\]
we will first decompose the integrand into partial fractions.
### Step 1: Partial Fraction Decomposition
We can express the integrand as:
\[
\frac{2x^2 + 3}{(x^2 - 1)(x^2 + 4)} = \frac{A}{x^2 - 1} + \frac{B}{x^2 + 4}.
\]
Multiplying through by the denominator \((x^2 - 1)(x^2 + 4)\), we have:
\[
2x^2 + 3 = A(x^2 + 4) + B(x^2 - 1).
\]
Expanding the right side:
\[
2x^2 + 3 = Ax^2 + 4A + Bx^2 - B = (A + B)x^2 + (4A - B).
\]
### Step 2: Set Up the System of Equations
By equating coefficients, we get the following system of equations:
1. \(A + B = 2\)
2. \(4A - B = 3\)
### Step 3: Solve the System of Equations
From the first equation, we can express \(B\) in terms of \(A\):
\[
B = 2 - A.
\]
Substituting this into the second equation:
\[
4A - (2 - A) = 3 \implies 4A - 2 + A = 3 \implies 5A = 5 \implies A = 1.
\]
Now substituting \(A = 1\) back into the first equation:
\[
1 + B = 2 \implies B = 1.
\]
### Step 4: Rewrite the Integral
Now we can rewrite the integral:
\[
\int \frac{2x^2 + 3}{(x^2 - 1)(x^2 + 4)} \, dx = \int \left( \frac{1}{x^2 - 1} + \frac{1}{x^2 + 4} \right) \, dx.
\]
### Step 5: Integrate Each Term
Now, we integrate each term separately:
1. For \(\int \frac{1}{x^2 - 1} \, dx\), we use the formula:
\[
\int \frac{1}{x^2 - a^2} \, dx = \frac{1}{2a} \log \left| \frac{x - a}{x + a} \right| + C.
\]
Here, \(a = 1\):
\[
\int \frac{1}{x^2 - 1} \, dx = \frac{1}{2} \log \left| \frac{x - 1}{x + 1} \right|.
\]
2. For \(\int \frac{1}{x^2 + 4} \, dx\), we use the formula:
\[
\int \frac{1}{x^2 + a^2} \, dx = \frac{1}{a} \tan^{-1} \left( \frac{x}{a} \right) + C.
\]
Here, \(a = 2\):
\[
\int \frac{1}{x^2 + 4} \, dx = \frac{1}{2} \tan^{-1} \left( \frac{x}{2} \right).
\]
### Step 6: Combine the Results
Combining both results, we have:
\[
\int \frac{2x^2 + 3}{(x^2 - 1)(x^2 + 4)} \, dx = \frac{1}{2} \log \left| \frac{x - 1}{x + 1} \right| + \frac{1}{2} \tan^{-1} \left( \frac{x}{2} \right) + C.
\]
### Step 7: Compare with Given Expression
We are given that:
\[
\int \frac{2x^2 + 3}{(x^2 - 1)(x^2 + 4)} \, dx = k \log \left( \frac{x + 1}{x - 1} \right) + k \tan^{-1} \left( \frac{x}{2} \right).
\]
From our result, we can see that:
\[
\frac{1}{2} \log \left| \frac{x - 1}{x + 1} \right| = -\frac{1}{2} \log \left( \frac{x + 1}{x - 1} \right).
\]
Thus, we can conclude that \(k = \frac{1}{2}\).
### Final Answer
Therefore, the value of \(k\) is:
\[
\boxed{\frac{1}{2}}.
\]
