## College Algebra 7th Edition

$x=\pm\sqrt{2}$
$1+\displaystyle \frac{1}{(x+1)(x+2)}=\frac{2}{x+1}+\frac{1}{x+2}$ We multiply through by $(x+1)(x+2)$: $(x+1)(x+2)+1=2(x+2)+(x+1)$ And distribute: $x^{2}+3x+2+1=2x+4+x+1$ And simplify: $x^{2}+3x+2+1-2x-4-x-1=0$ $x^{2}-2=0$ $x=\pm\sqrt{2}$