We are asked to simplify the expression $\frac{x}{x+1} - \frac{3x}{x^2-1}$.

1. Problem Description

We are asked to simplify the expression $\frac{x}{x+1} - \frac{3x}{x^2-1}$.

2. Solution Steps

First, we factor the denominator of the second term:

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

Thus, the expression becomes:

$\frac{x}{x+1} - \frac{3x}{(x+1)(x-1)}$

To subtract these fractions, we need a common denominator. The least common denominator is $(x+1)(x-1)$. We multiply the first term by $\frac{x-1}{x-1}$ to get a common denominator:

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

Now we can rewrite the expression as:

$\frac{x^2-x}{(x+1)(x-1)} - \frac{3x}{(x+1)(x-1)}$

Now we can subtract the numerators:

$\frac{x^2-x-3x}{(x+1)(x-1)} = \frac{x^2-4x}{(x+1)(x-1)}$

Next, we factor the numerator:

$x^2 - 4x = x(x-4)$

So the expression becomes:

$\frac{x(x-4)}{(x+1)(x-1)}$

Therefore, the simplified expression is $\frac{x(x-4)}{(x+1)(x-1)}$.

3. Final Answer

$\frac{x(x-4)}{(x+1)(x-1)}$