To solve the inequality \(\frac{(x-3)(x+2)}{x^2-1} < 1\), we will follow these steps:
### Step 1: Rearranging the Inequality
First, we will move 1 to the left side of the inequality:
\[
\frac{(x-3)(x+2)}{x^2-1} - 1 < 0
\]
### Step 2: Finding a Common Denominator
To combine the terms, we will express 1 with the common denominator \(x^2 - 1\):
\[
\frac{(x-3)(x+2)}{x^2-1} - \frac{x^2 - 1}{x^2 - 1} < 0
\]
This simplifies to:
\[
\frac{(x-3)(x+2) - (x^2 - 1)}{x^2 - 1} < 0
\]
### Step 3: Simplifying the Numerator
Now we simplify the numerator:
\[
(x-3)(x+2) = x^2 - 3x + 2x - 6 = x^2 - x - 6
\]
So, the numerator becomes:
\[
x^2 - x - 6 - (x^2 - 1) = -x - 6 + 1 = -x - 5
\]
Thus, we have:
\[
\frac{-x - 5}{x^2 - 1} < 0
\]
### Step 4: Factoring the Denominator
The denominator can be factored as:
\[
x^2 - 1 = (x - 1)(x + 1)
\]
So the inequality now looks like:
\[
\frac{-x - 5}{(x - 1)(x + 1)} < 0
\]
### Step 5: Finding Critical Points
Next, we find the critical points by setting the numerator and denominator to zero:
1. Numerator: \(-x - 5 = 0 \Rightarrow x = -5\)
2. Denominator: \(x - 1 = 0 \Rightarrow x = 1\) and \(x + 1 = 0 \Rightarrow x = -1\)
The critical points are \(x = -5\), \(x = -1\), and \(x = 1\).
### Step 6: Testing Intervals
We will test the intervals created by these critical points:
- \( (-\infty, -5) \)
- \( (-5, -1) \)
- \( (-1, 1) \)
- \( (1, \infty) \)
1. **Interval \((- \infty, -5)\)**: Choose \(x = -6\)
\[
\frac{-(-6) - 5}{(-6 - 1)(-6 + 1)} = \frac{6 - 5}{(-7)(-5)} = \frac{1}{35} > 0
\]
2. **Interval \((-5, -1)\)**: Choose \(x = -3\)
\[
\frac{-(-3) - 5}{(-3 - 1)(-3 + 1)} = \frac{3 - 5}{(-4)(-2)} = \frac{-2}{8} < 0
\]
3. **Interval \((-1, 1)\)**: Choose \(x = 0\)
\[
\frac{-(0) - 5}{(0 - 1)(0 + 1)} = \frac{-5}{(-1)(1)} = 5 > 0
\]
4. **Interval \((1, \infty)\)**: Choose \(x = 2\)
\[
\frac{-(2) - 5}{(2 - 1)(2 + 1)} = \frac{-2 - 5}{(1)(3)} = \frac{-7}{3} < 0
\]
### Step 7: Conclusion
The inequality \(\frac{-x - 5}{(x - 1)(x + 1)} < 0\) holds in the intervals \((-5, -1)\) and \((1, \infty)\). Therefore, the solution set is:
\[
x \in (-5, -1) \cup (1, \infty)
\]
