Solve the equation $\frac{a+1}{a(x+3)} - \frac{3ax}{a(x^2+2x-3)} = \frac{4}{x-1}$.

1. Problem Description

Solve the equation $\frac{a+1}{a(x+3)} - \frac{3ax}{a(x^2+2x-3)} = \frac{4}{x-1}$.

2. Solution Steps

First, we simplify the expression $x^2+2x-3$ by factoring it:

$x^2+2x-3 = (x+3)(x-1)$

Now, substitute this back into the original equation:

$\frac{a+1}{a(x+3)} - \frac{3ax}{a(x+3)(x-1)} = \frac{4}{x-1}$

Multiply both sides of the equation by $a(x+3)(x-1)$ to eliminate the denominators:

$(a+1)(x-1) - 3ax = 4a(x+3)$

$ax - a + x - 1 - 3ax = 4ax + 12a$

$-2ax - a + x - 1 = 4ax + 12a$

Combine like terms:

$x - 1 = 6ax + 13a$

We are looking for a solution in terms of $x$. Isolate $x$ terms on one side:

$x - 6ax = 13a + 1$

$x(1 - 6a) = 13a + 1$

Divide both sides by $(1 - 6a)$ to solve for $x$:

$x = \frac{13a + 1}{1 - 6a}$

3. Final Answer

$x = \frac{13a + 1}{1 - 6a}$