Solve `(x+2)/(x^2+1)gt 1/2 `

Explanation

To solve the inequality \(\frac{x+2}{x^2+1} > \frac{1}{2}\), we will follow these steps:

### Step 1: Rewrite the Inequality
We start by rewriting the inequality:
\[ \frac{x+2}{x^2+1} - \frac{1}{2} > 0 \]

### Step 2: Find a Common Denominator
To combine the fractions, we need a common denominator. The common denominator is \(2(x^2 + 1)\):
\[ \frac{2(x + 2) - (x^2 + 1)}{2(x^2 + 1)} > 0 \]

### Step 3: Simplify the Numerator
Now, simplify the numerator:
\[ 2(x + 2) - (x^2 + 1) = 2x + 4 - x^2 - 1 = -x^2 + 2x + 3 \]
Thus, the inequality becomes:
\[ \frac{-x^2 + 2x + 3}{2(x^2 + 1)} > 0 \]

### Step 4: Analyze the Denominator
The denominator \(2(x^2 + 1)\) is always positive for all real \(x\) since \(x^2 + 1 > 0\). Therefore, we only need to focus on the numerator:
\[ -x^2 + 2x + 3 > 0 \]

### Step 5: Rearrange the Inequality
Rearranging gives us:
\[ -x^2 + 2x + 3 > 0 \implies x^2 - 2x - 3 < 0 \]

### Step 6: Factor the Quadratic
Next, we factor the quadratic:
\[ x^2 - 2x - 3 = (x - 3)(x + 1) \]
Thus, the inequality becomes:
\[ (x - 3)(x + 1) < 0 \]

### Step 7: Determine the Critical Points
The critical points are \(x = -1\) and \(x = 3\). We will test intervals around these points to determine where the product is negative.

### Step 8: Test the Intervals
1. **Interval \((-∞, -1)\)**: Choose \(x = -2\):
\[ (-2 - 3)(-2 + 1) = (-5)(-1) = 5 > 0 \]
2. **Interval \((-1, 3)\)**: Choose \(x = 0\):
\[ (0 - 3)(0 + 1) = (-3)(1) = -3 < 0 \]
3. **Interval \((3, ∞)\)**: Choose \(x = 4\):
\[ (4 - 3)(4 + 1) = (1)(5) = 5 > 0 \]

### Step 9: Conclusion
The product \((x - 3)(x + 1) < 0\) is satisfied in the interval:
\[ (-1, 3) \]

Thus, the solution to the inequality \(\frac{x+2}{x^2+1} > \frac{1}{2}\) is:
\[ \boxed{(-1, 3)} \]