I want to show that the bilinear form $$a(u,v):=\int_{0}^{1} (-(1+x^2)u'v'-3xu'v) ~\text{d}x$$ is coercive in $H_{0}^1(0,1)$, in order to solve the problem $$(1+x^2)u''-xu'=\sin(2 \pi x)$$ with zero boundary values, so I have:
$$a(u,u)=\int_{0}^{1} (-(1+x^2)(u')^2 -3xu' u )~\text{d}x=\int_{0}^{1} (-(1+x^2)(u')^2+\dfrac{3}{2}u^2)~\text{d}x$$
but I'm stuck in obtaining $0<\theta$ such that $$\theta |u|_{H_{0}^{1}}^2=\theta \int u^2+(u')^2 ~\text{d}x\leq a(u,u)$$ (coercive), and in fact I believe is not possible, because the leading term is "so negative!". Any suggestion even in the formulation would be appreciated.