Step 1: Understanding the Concept:
This is an integration problem requiring partial fraction decomposition.
The denominator has a linear factor $(x-1)$ and an irreducible quadratic factor $(x^2+1)$.
After integration, we need to compare the result with the given expression to find the value of the exponent '$a$'.
Step 2: Key Formula or Approach:
Set up partial fractions: $\frac{2x+3}{(x-1)(x^2+1)} = \frac{B}{x-1} + \frac{Cx+D}{x^2+1}$.
Solve for constants $B, C, D$.
Integrate each term separately.
Use log properties like $n \ln x = \ln(x^n)$ and $\ln x + \ln y = \ln(xy)$ to combine terms.
Step 3: Detailed Explanation:
Let's decompose the integrand into partial fractions:
\[ \frac{2x+3}{(x-1)(x^2+1)} = \frac{B}{x-1} + \frac{Cx+D}{x^2+1} \]
Multiply through by the common denominator $(x-1)(x^2+1)$:
\[ 2x + 3 = B(x^2 + 1) + (Cx + D)(x - 1) \]
To find $B$, let $x = 1$:
\[ 2(1) + 3 = B(1^2 + 1) + 0 \Rightarrow 5 = 2B \Rightarrow B = \frac{5}{2} \]
To find $D$, let $x = 0$:
\[ 3 = B(1) + (D)(-1) \Rightarrow 3 = \frac{5}{2} - D \Rightarrow D = \frac{5}{2} - 3 = -\frac{1}{2} \]
To find $C$, compare the coefficients of $x^2$ on both sides:
\[ 0 = B + C \Rightarrow C = -B = -\frac{5}{2} \]
Now, substitute the partial fractions back into the integral:
\[ \int \frac{2x+3}{(x-1)(x^2+1)} dx = \int \left( \frac{\frac{5}{2}}{x-1} + \frac{-\frac{5}{2}x - \frac{1}{2}}{x^2+1} \right) dx \]
Split the integral:
\[ = \frac{5}{2} \int \frac{1}{x-1} dx - \frac{5}{2} \int \frac{x}{x^2+1} dx - \frac{1}{2} \int \frac{1}{x^2+1} dx \]
For the second integral, multiply and divide by 2:
\[ = \frac{5}{2} \ln|x-1| - \frac{5}{4} \int \frac{2x}{x^2+1} dx - \frac{1}{2} \tan^{-1} x \]
\[ = \frac{5}{2} \ln|x-1| - \frac{5}{4} \ln|x^2+1| - \frac{1}{2} \tan^{-1} x + C_1 \]
Combine the log terms using properties of logarithms:
\[ = \ln(|x-1|^{5/2}) + \ln((x^2+1)^{-5/4}) - \frac{1}{2} \tan^{-1} x + C_1 \]
\[ = \ln \left( |x-1|^{5/2} (x^2+1)^{-5/4} \right) - \frac{1}{2} \tan^{-1} x + C_1 \]
Comparing this result with the given expression $\log_e \left\{ (x - 1)^{\frac{5}{2}} (x^2 + 1)^a \right\} - \frac{1}{2} \tan^{-1} x + A$, we can see that:
\[ a = -\frac{5}{4} \]
Step 4: Final Answer:
The value of $a$ is $-\frac{5}{4}$.