To simplify the expression, we first need to find a common denominator for the fractions on the left side of the equation.

The common denominator for ( x-1 ) and ( x+1 ) is ( (x-1)(x+1) = x^2 - 1 ).

So, the left side of the equation becomes:

[ \frac{3(x+1)}{x^2-1} - \frac{5(x-1)}{x^2-1} - 8 = \frac{13}{1-x^2} ]

Now we can simplify further:

[ \frac{3x+3-5x+5-8(x^2-1)}{x^2-1} = \frac{13}{1-x^2} ]
[ \frac{3x+3-5x+5-8x^2+8}{x^2-1} = \frac{13}{1-x^2} ]
[ \frac{-8x^2 - 2x + 16}{x^2-1} = \frac{13}{1-x^2} ]

As both denominators are now the same, we can equate the numerators:

[ -8x^2 - 2x + 16 = 13 ]

Now, we solve for ( x ):

[ -8x^2 - 2x + 16 = 13 ]
[ -8x^2 - 2x + 3 = 0 ]

Now, we can either factor or use the quadratic formula to find the solutions for ( x ). Let's use the quadratic formula:

[ x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(-8)(3)}}{2(-8)} ]
[ x = \frac{2 \pm \sqrt{4 + 96}}{-16} ]
[ x = \frac{2 \pm \sqrt{100}}{-16} ]
[ x = \frac{2 \pm 10}{-16} ]

Therefore, the solutions for ( x ) are:

[ x = \frac{2 + 10}{-16} = \frac{12}{-16} = -\frac{3}{4} ]

or

[ x = \frac{2 - 10}{-16} = \frac{-8}{-16} = \frac{1}{2} ]