First you want to expand both of the quadratic denominators. To do this for #x^2+4x-5# which is in standard form, you would need two factors of #-5# that add together to get #+4#. That would be the factors #5# and #-1#. So it expands to #(x+5)(x-1)#.

Now we have #4/((x+5)(x-1)) - 3/(x^2-1)#

You might recognize #x^2-1# as a perfect square that expands out to #(x+1)(x-1)#.

Now we have #4/((x+5)(x-1)) - 3/((x+1)(x-1))#

In order to add or subtract fractions we need to have like denominators. So in this case we can do some manipulation to meet that criterion like so:

First we can multiply #4/((x+5)(x-1))# by the factor it's missing in its denominator which would be #(x+1)#.

#4/((x+5)(x-1))*(x+1)/(x+1) = (4(x+1))/((x+5)(x-1)(x+1)) = (4x+4)/((x+5)(x-1)(x+1))#

Next, we should multiply #3/((x+1)(x-1))# by the factor it's missing in its denominator which would be #(x+5)#.

#3/((x+1)(x-1)) * (x+5)/(x+5) = (3(x+5))/((x+5)(x-1)(x+1)) = (3x+15)/((x+5)(x-1)(x+1)#

By multiplying each fraction by a factor over itself it is the same as multiplying by one. Finally, we have the same denominator for each fraction and we can complete the subtraction in the numerator while, of course, the denominator will stay the same.

#((4x+4)-(3x+15))/((x+5)(x-1)(x+1)) = (4x+4-3x-15)/((x+5)(x-1)(x+1)) = (x-11)/((x+5)(x-1)(x+1))#