How to use the initial condition in this DE: $x^2\frac{dy}{dx}=\frac{4x^2-x-2}{\left(x+1\right)\left(y+1\right)}$ Here is my proposed solution:
\begin{align}
x^2\frac{dy}{dx}&=\frac{4x^2-x-2}{\left(x+1\right)\left(y+1\right)}\tag{1}\\
\implies & \left(y+1\right)\frac{dy}{dx}=\frac{4x^2-x-2}{x^2\left(x+1\right)}\tag{2}\\
\implies & \int\frac{dy}{dx}\left(y+1\right)\:dx=\int\frac{4x^2-x-2}{x^2\left(x+1\right)}\:dx\tag{3}\\
\implies & \frac{y^2}{2}+y=\int\alpha\left(x\right)\:dx,\tag{4}
\end{align}
and here I'm letting $\alpha\left(x\right)$ denote the partial fraction decomposition of the integrand in the RHS, which is
\begin{align}
\alpha\left(x\right)=\frac{4x^2-x-2}{x^2\left(x+1\right)}=\frac{1}{x}-\frac{2}{x^2}+\frac{3}{x+1},\tag{5}
\end{align}
and this gives me
\begin{align}
\frac{y^2}{2}+y=\log\left|x\right|+\frac{2}{x}+3\log\left|x+1\right|+C.\tag{6}
\end{align}
Okay, so no problems there. But now the IC is such that $y\left(1\right)=1.$ I don't know what to do with these two numbers because this is an implicit answer and I cannot separate the $y$'s.
Thank you for your time,

$y(1)=1$ implies (by plugging 1 into both $x$ and $y$)
$$\frac12+1=\log 1+\frac21+3\log 2+C.$$
Now, solve for $C$.
