Question 42080
{{{3/(x^2 + 4x + 3)}}}
= {{{3/(x^2 + x + 3x + 3)}}}
= {{{3/(x(x + 1) + 3(x + 1))}}}
= {{{3/((x+1)(x+3))}}}



{{{1/(x^2-9)}}}
= {{{1/((x+3)(x-3))}}}



Hence, {{{3/(x^2 + 4x + 3) - 1/(x^2-9)}}}
= {{{3/((x+1)(x+3)) - 1/((x+3)(x-3))}}}
= {{{(3(x-3) - 1(x+1))/((x+1)(x+3)(x-3))}}}
= {{{(2x - 10)/((x+1)(x+3)(x-3))}}}
= {{{2(x-5)/((x+1)(x+3)(x-3))}}}