Answer
$\dfrac{4}{x^{2}+4x+3}+\dfrac{3}{x^{2}-x-2}=\dfrac{7x+1}{(x+1)(x+3)(x-2)}$
Work Step by Step
$\dfrac{4}{x^{2}+4x+3}+\dfrac{3}{x^{2}-x-2}$
Factor the denominators of both fractions:
$\dfrac{4}{x^{2}+4x+3}+\dfrac{3}{x^{2}-x-2}=...$
$...=\dfrac{4}{(x+3)(x+1)}+\dfrac{3}{(x-2)(x+1)}=...$
Evaluate the sum of these two fractions using the LCD, which is $(x+1)(x+3)(x-2)$ in this case:
$...=\dfrac{4(x-2)+3(x+3)}{(x+1)(x+3)(x-2)}=...$
Simplify:
$...=\dfrac{4x-8+3x+9}{(x+1)(x+3)(x-2)}=...$
$...=\dfrac{7x+1}{(x+1)(x+3)(x-2)}$
