Answer
$y=-\dfrac{2}{(x+1)^2+C}$
Work Step by Step
We are given that $\dfrac{dy}{dx}=(x+1)y^2$
We will separate the variables to obtain:
$\dfrac{dy}{y^2}=(x+1) \ dx$
Integrate to obtain:
$\int \dfrac{dy}{y^2}=\int (x+1) dx$
This implies that $y^{-2} \ dy=\int (x+1) \ dx$
or, $-\dfrac{1}{y}=\dfrac{(x+1)^2}{2}+C$
Therefore, we have: $y=-\dfrac{2}{(x+1)^2+C}$
