Answer
$x=3$
Work Step by Step
$\frac{1}{x+2}+\frac{4}{x^2-4}=1$
$\frac{1}{x+2}+\frac{4}{(x-2)(x+2)}=1$
$\frac{1(x-2)}{(x+2)(x-2)}+\frac{4}{(x-2)(x+2)}=\frac{(x-2)(x+2)}{(x-2)(x+2)}$
$x-2+4=(x-2)(x+2)$
$x+2=x^2-4$
$x^2-x-6=0$
$(x-3)(x+2)=0$
$x-3=0$
$x=3$
$x+2=0$
$x=-2$ (invalid answer since the denominator would be zero)
$x=3$
$\frac{1}{x+2}+\frac{4}{x^2-4}=1$
$\frac{1}{3+2}+\frac{4}{3^2-4}=1$
$1/5 + 4/(9-4) = 1$
$1/5 + 4/5 = 1$
