Solve` (5x+1)/((x+1)^2)lt 1`

Explanation

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

### Step 1: Rearranging the Inequality
Start by subtracting 1 from both sides of the inequality:
\[ \frac{5x + 1}{(x + 1)^2} - 1 < 0 \]
This simplifies to:
\[ \frac{5x + 1 - (x + 1)^2}{(x + 1)^2} < 0 \]

### Step 2: Simplifying the Numerator
Now, simplify the numerator:
\[ 5x + 1 - (x^2 + 2x + 1) < 0 \]
This expands to:
\[ 5x + 1 - x^2 - 2x - 1 < 0 \]
Combining like terms gives:
\[ -x^2 + 3x < 0 \]
or
\[ -x^2 + 3x = -(x^2 - 3x) < 0 \]

### Step 3: Factoring the Expression
Factor out the negative sign:
\[ -(x(x - 3)) < 0 \]
This implies:
\[ x(x - 3) > 0 \]

### Step 4: Finding Critical Points
The critical points where the expression equals zero are:
\[ x = 0 \quad \text{and} \quad x = 3 \]

### Step 5: Analyzing Intervals
Now we will analyze the sign of \(x(x - 3)\) in the intervals determined by the critical points:
1. \( (-\infty, 0) \)
2. \( (0, 3) \)
3. \( (3, \infty) \)

- **Interval \( (-\infty, 0) \)**: Choose \(x = -1\):
\[ (-1)(-1 - 3) = (-1)(-4) = 4 > 0 \quad \text{(Positive)} \]

- **Interval \( (0, 3) \)**: Choose \(x = 1\):
\[ (1)(1 - 3) = (1)(-2) = -2 < 0 \quad \text{(Negative)} \]

- **Interval \( (3, \infty) \)**: Choose \(x = 4\):
\[ (4)(4 - 3) = (4)(1) = 4 > 0 \quad \text{(Positive)} \]

### Step 6: Conclusion on Intervals
The expression \(x(x - 3) > 0\) is satisfied in the intervals:
\[ (-\infty, 0) \cup (3, \infty) \]

### Step 7: Exclusion of Points
However, we must also consider that the original inequality has a denominator \((x + 1)^2\) which is not defined at \(x = -1\). Therefore, we exclude this point from our solution set.

### Final Solution
The solution to the inequality \(\frac{5x + 1}{(x + 1)^2} < 1\) is:
\[ x \in (-\infty, -1) \cup (3, \infty) \]