First order differential equation integrating factor is $e^{\int\frac{2}{x^2-1}}$ So i got the first order ode
$$(x^2-1)\frac{dy}{dx}+2xy=x$$
I divided both sides by $x^2-1$
$$\frac{dy}{dx}+\frac{2}{x^2-1}xy=\frac{x}{x^2-1}$$
in the form $y' + p(x)y = q(x)$
So that means the integrand is...
$$e^{\int\frac{2}{x^2-1}}$$
But i'm not sure what to do i think the $\int\frac{2}{x^2-1}$ = $-\log{(x-1)}+4\log{(x+1)}$
So it's $$e^{-\log{(x-1)}+4\log{(x+1)}}$$
$$e^{\log{(x-1)^{-1}}+\log{(x+1)^4}}$$
$$\frac{1}{x-1}+(x+1)^4$$
Is this right? and then just multiply both sides by this?

I do not know if you have been told to use the the integration factor, but you can make it a lot easier for yourself. Rearrange for $\frac{dy}{dx}$:
\begin{equation*}
\frac{dy}{dx}=-\frac{x(2y-1)}{x^2-1}.
\end{equation*}
Divide both sides by $2y-1$and integrate with respect to $x$ gives
\begin{equation*}
\frac{1}{2}\ln(2y-1)=-\frac{1}{2}\ln (x^2-1)+C
\end{equation*}
where $C$ is a constant. Solving for $y$ gives the result.
