Answer
$\ln|y|=\dfrac{-1}{2(x^2+1)}+\dfrac{1}{2}$
Work Step by Step
We are given that $ \dfrac{dy}{dx}=\dfrac{xy}{(x^2+1)^2}$
We will separate the variables to obtain:
$\dfrac{dy}{y}=\dfrac{x \ dx}{(x^2+1)^2}$
Integrate to obtain:
$\int \dfrac{dy}{y}=\int \dfrac{x \ dx}{(x^2+1)^2} \ dx$
This implies that $\ln|y|=\dfrac{-1}{2(x^2+1)}+C$
After applying the initial conditions, $y=1$ when $x=0$, we get $C=\dfrac{1}{2}$
Therefore, we have: $\ln|y|=\dfrac{-1}{2(x^2+1)}+\dfrac{1}{2}$
