Qulve por $$ x $$ in the equation $$ \frac{x+2}{x-1}-\frac{x-3}{x+1}=\frac{4 x}{x^{2}-1} $$

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We start with the equation
$$
\frac{x+2}{x-1} - \frac{x-3}{x+1} = \frac{4x}{x^2-1}.
$$
Notice that
$$
x^2-1=(x-1)(x+1).
$$
We must have $$ x \neq 1 $$ and $$ x \neq -1 $$ to avoid division by zero.

Multiply both sides of the equation by $$(x-1)(x+1)$$ to eliminate the denominators:
$$
(x+2)(x+1) - (x-3)(x-1) = 4x.
$$

Next, we expand the products:
$$
(x+2)(x+1)= x^2 + 3x + 2,
$$
$$
(x-3)(x-1)= x^2 - 4x + 3.
$$

Now, subtract the second expression from the first:
$$
(x^2 + 3x + 2) - (x^2 - 4x + 3) = x^2 + 3x + 2 - x^2 + 4x - 3.
$$
The $$ x^2 $$ terms cancel, leaving:
$$
7x - 1.
$$

So, the equation becomes:
$$
7x - 1 = 4x.
$$

Subtract $$ 4x $$ from both sides:
$$
7x - 4x - 1 = 0,
$$
$$
3x - 1 = 0.
$$

Finally, solve for $$ x $$ by adding 1 to both sides and dividing by 3:
$$
3x = 1 \quad \Longrightarrow \quad x = \frac{1}{3}.
$$

Since $$ \frac{1}{3} $$ does not equal $$ 1 $$ or $$-1$$, it is valid. Thus, the solution is:
$$
x = \frac{1}{3}.
$$
$$ x = \frac{1}{3} $$

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