Step 1: Understanding the Concept:
The integrand can be decomposed into partial fractions. Since only \(x^2\) terms are present, we can use a temporary substitution to simplify the algebra.
Step 2: Key Formula or Approach:
1. Partial fraction decomposition.
2. Integration formula: \(\int \frac{1}{x^2-a^2} dx = \frac{1}{2a} \log |\frac{x-a}{x+a}| + c\).
Step 3: Detailed Explanation:
Let \(x^2 = t\) in the expression \(\frac{2x^2+3}{(x^2-1)(x^2-4)}\).
\[ \frac{2t+3}{(t-1)(t-4)} = \frac{A}{t-1} + \frac{B}{t-4} \]
Using the "cover-up" method:
\(A = \frac{2(1)+3}{1-4} = -\frac{5}{3}\) and \(B = \frac{2(4)+3}{4-1} = \frac{11}{3}\).
The integral becomes:
\[ I = \int [ -\frac{5/3}{x^2-1} + \frac{11/3}{x^2-4} ] dx \]
\[ I = -\frac{5}{3} \cdot \frac{1}{2(1)} \log |\frac{x-1}{x+1}| + \frac{11}{3} \cdot \frac{1}{2(2)} \log |\frac{x-2}{x+2}| \]
\[ I = \frac{5}{6} \log |\frac{x+1}{x-1}| + \frac{11}{12} \log |\frac{x-2}{x+2}| \]
To combine, use the same denominator \(12\):
\[ I = \log | \left( \frac{x-2}{x+2} \right)^{11/12} \cdot \left( \frac{x+1}{x-1} \right)^{10/12} | + c \]
Comparing with the given form, \(a = \frac{11}{12}\) and \(b = \frac{10}{12}\).
Then, \(a+b = \frac{11}{12} + \frac{10}{12} = \frac{21}{12}\).
Step 4: Final Answer:
The value of \(a+b\) is \(\frac{21}{12}\).