Answer :
[tex]D:x+2\neq0\to x\neq-2\ and\ x-3\neq0\to x\neq3\\\\\frac{3}{x+2}+\frac{4}{x-3}=\frac{3(x-3)}{(x+2)(x-3)}+\frac{4(x+2)}{(x+2)(x-3)}=\frac{3(x)+3(-3)+4(x)+4(2)}{(x)(x)+(x)(-3)+(2)(x)+2(-3)}\\\\=\frac{3x-9+4x+8}{x^2-3x+2x-6}=\boxed{\frac{7x-1}{x^2-x-6}}[/tex]
[tex]\frac{3}{x + 2} + \frac{4}{x - 3} = \frac{3(x - 3)}{(x + 2)(x - 3)} + \frac{4(x + 2)}{(x + 2)(x - 3)} = \frac{3x - 9}{x^{2} - x - 6} + \frac{4x+ 8}{x^{2} - x - 6} =\frac{7x -1}{x^{2} - x - 6} [/tex]
