Divide everything by $(1-x^2)$ to get $y''$ by itself.
$$y'' - {2x\over {1 - x^2}}y' + {α(α + 1)\over {1 - x^2}}y = 0$$
Now that it is in the proper form, we can use Abel's theorem of $W = ce^{\int-p(x)dx}$ where c is a constant and $p(x)$ is $- {2x\over {1 - x^2}}$ in this case. Now we solve the integral:
$$ce^{\int{2x\over {1 - x^2}}dx}$$
Using the substitution $u = 1-x^2$ and $du =-2xdx$ we get
$$ ce^{-\int{1\over u}du} = ce^{-ln(u)+C} = ce^{-ln(1-x^2)} = c(1-x^2)^{-1}e^C = {ce^C\over {1-x^2}}$$
But $ce^C$ is just some constant, so we can subsume it into just $c$. Simplifying this, we get that the Wronskian is:
$$W = {c\over {1-x^2}}$$
