Question 146188
Essentially what you have here is

{{{(x+3)/(x+1)*(1/(x-1))-2/(x-1)=(x+3)/((x+1)(x-1))-2/(x-1)}}}

Now, what is the {{{-2/(x-1)}}} term missing from the denominator????
(x+1).

Thus,

={{{(x+3)/((x+1)(x-1))-2(x+1)/((x-1)(x+1))}}}

At this point we can combine the numerators.

={{{(x+3-2x-2)/((x+1)(x-1))}}}
={{{(-x+1)/((x+1)(x-1))}}}
={{{-(x-1)/((x+1)(x-1))}}}
={{{-1/(x+1)}}}

And that would be that.