Question 971105
{{{x/(x^2-9)-(x+1)/(x^2+6x+9)}}}


1).  Find the LCD of both fractions:  {{{x^2-9}}} can be factored as {{{(x+3)(x-3)}}}
                                                        {{{X^2+6x+9}}} can be factored as {{{(x+3)^2}}}


2).  Replace the denominators of each fraction with {{{(x+3)^2*(x-3)}}}:


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


3).  Multiply the numerator of fraction #1 by {{{x+3}}}:


{{{(x(x+3))/((x+3)^2*(x-3))}}} -----> {{{(x^2+3x)/((x+3)^2*(x-3))}}}


4).  Multiply the numerator of fraction #2 by {{{x-3}}}:


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


5).  Subtract numerator of fraction #2 from fraction #1 numerator:


{{{(x^2+3x-x^2+2x+3)/((x+3)^2*(x-3))}}} -----> {{{(5x+3)/((x+3)^2*(x-3))}}}


Final Answer:  {{{(5x+3)/((x+3)^2*(x-3))}}}