Answer
$x=-\frac{1}{2}$ or $x=-2$
Work Step by Step
$\displaystyle \frac{1}{(x+1)^{2}}=\frac{1}{x+1}+2$
We multiply both sides by $(x+1)^{2}$:
$1=(x+1)+2(x+1)^{2}$
$2(x+1)^{2}+(x+1)-1=0$
We solve by factoring:
$[2(x+1)-1][(x+1)+1]=0$
$(2x+2-1)(x+2)=0$
$(2x+1)(x+2)=0$
$(2x+1)=0$ or $(x+2)=0$
$x=-\frac{1}{2}$ or $x=-2$