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?

your equation $$(x^2 - 1) \frac{dy}{dx} + 2x y = x $$ is an exact differential equation. the reason is it can be written as $$\frac d{dx}\left((x^2 - 1) y\right) = x $$ on integration gives you $$(x^2 - 1)y = \frac12 x^2 + c \to y = \frac{x^2}{2(x^2 - 1)} + \frac c{x^2 - 1}$$
