The given expression is:

$\frac{1}{x^2 + 2x + 1} - \frac{1}{x^2 - 1}.$

Step 1: Factorize the denominators

1. $x^2 + 2x + 1 = (x+1)^2$,
2. $x^2 - 1 = (x-1)(x+1)$.

Thus, the expression becomes:

$\frac{1}{(x+1)^2} - \frac{1}{(x-1)(x+1)}.$

Step 2: Find the common denominator

The least common denominator (LCD) is $(x+1)^2(x-1)$.

Step 3: Rewrite the fractions with the common denominator

1. Rewrite $\frac{1}{(x+1)^2}$ with the LCD: $\frac{1}{(x+1)^2} = \frac{1 \cdot (x-1)}{(x+1)^2(x-1)} = \frac{x-1}{(x+1)^2(x-1)}.$

2. Rewrite $\frac{1}{(x-1)(x+1)}$ with the LCD: $\frac{1}{(x-1)(x+1)} = \frac{1 \cdot (x+1)}{(x+1)^2(x-1)} = \frac{x+1}{(x+1)^2(x-1)}.$

Now the expression becomes: $\frac{x-1}{(x+1)^2(x-1)} - \frac{x+1}{(x+1)^2(x-1)}.$

Step 4: Combine the fractions

Combine the numerators over the common denominator: $\frac{(x-1) - (x+1)}{(x+1)^2(x-1)}.$

Simplify the numerator: $(x-1) - (x+1) = x - 1 - x - 1 = -2.$

Thus, the expression simplifies to: $\frac{-2}{(x+1)^2(x-1)}.$

Final Answer:

$\boxed{\frac{-2}{(x+1)^2(x-1)}}.$

Would you like a detailed explanation of any step or to expand on related topics?

Related Questions:

1. How do you simplify rational expressions involving polynomials?
2. What is the process of finding the least common denominator for rational expressions?
3. How can you determine restrictions for the variable in rational expressions?
4. How does factoring help in simplifying algebraic fractions?
5. What are the differences between rational and polynomial expressions?

Tip:

Always check for restrictions (e.g., $x+1 \neq 0$ and $x-1 \neq 0$) to ensure the expression is well-defined.