Solve ` (5x+1)/((x+1)^2)` - 1 < 0

Explanation

To solve the inequality \(\frac{5x + 1}{(x + 1)^2} - 1 < 0\), we will follow these steps:

### Step 1: Rewrite the Inequality
Start by rewriting the inequality:
\[ \frac{5x + 1}{(x + 1)^2} - 1 < 0 \]
This can be rewritten as:
\[ \frac{5x + 1 - (x + 1)^2}{(x + 1)^2} < 0 \]

### Step 2: Expand the Expression
Now, expand \((x + 1)^2\):
\[ (x + 1)^2 = x^2 + 2x + 1 \]
Substituting this back into the inequality gives:
\[ \frac{5x + 1 - (x^2 + 2x + 1)}{(x + 1)^2} < 0 \]

### Step 3: Simplify the Numerator
Now simplify the numerator:
\[ 5x + 1 - x^2 - 2x - 1 = -x^2 + 3x \]
Thus, the inequality becomes:
\[ \frac{-x^2 + 3x}{(x + 1)^2} < 0 \]

### Step 4: Factor the Numerator
Factor out the negative sign from the numerator:
\[ \frac{-x(x - 3)}{(x + 1)^2} < 0 \]
This can be rewritten as:
\[ \frac{x(x - 3)}{(x + 1)^2} > 0 \]

### Step 5: Identify Critical Points
Now, identify the critical points where the expression is zero or undefined:
- The numerator \(x(x - 3) = 0\) gives \(x = 0\) and \(x = 3\).
- The denominator \((x + 1)^2 = 0\) gives \(x = -1\) (undefined).

### Step 6: Test Intervals
Now we will test the intervals determined by the critical points:
- \((-∞, -1)\)
- \((-1, 0)\)
- \((0, 3)\)
- \((3, ∞)\)

1. **Interval \((-∞, -1)\)**: Choose \(x = -2\):
\[ \frac{-2(-2 - 3)}{(-2 + 1)^2} = \frac{-2 \cdot -5}{1} = 10 > 0 \]
2. **Interval \((-1, 0)\)**: Choose \(x = -0.5\):
\[ \frac{-0.5(-0.5 - 3)}{(-0.5 + 1)^2} = \frac{-0.5 \cdot -3.5}{0.25} = 7 > 0 \]
3. **Interval \((0, 3)\)**: Choose \(x = 1\):
\[ \frac{1(1 - 3)}{(1 + 1)^2} = \frac{1 \cdot -2}{4} = -0.5 < 0 \]
4. **Interval \((3, ∞)\)**: Choose \(x = 4\):
\[ \frac{4(4 - 3)}{(4 + 1)^2} = \frac{4 \cdot 1}{25} = 0.16 > 0 \]

### Step 7: Combine Results
The expression is positive in the intervals \((-∞, -1)\), \((-1, 0)\), and \((3, ∞)\), and negative in the interval \((0, 3)\).

### Step 8: Write the Solution
Since we are looking for where the expression is greater than zero:
\[ x \in (-\infty, -1) \cup (-1, 0) \cup (3, \infty) \]