I am dealing with the derivative which is part of a basic weak formula (with the weighted $v$ and solution $u$): $$\int_{0}^{1}\left(-v(x+2)\frac{d^2u}{dx^2}+vu\right)\,dx$$ $$\Big(\frac{du}{dx}\Big)\Big|_{x=0}=3$$$$ u(1)=2$$
I know that I will split it into two integrals (because of the x+2), but my problem is knowing how to deal with $\frac{d^2u}{dx^2}$.
This is what I tried for the second part (with the 2): $$f = -2v, df = -2v', dg = \frac{d^2u}{dx^2}, g = x\frac {du}{dx}=xu'$$ $$fg - \int_{0}^{1}g \,df =-2vxu'+\int_{0}^{1}2xu'v \,dx $$
This doesn't look good; I have the answer and there are no x's involved in the fg part at the beginning. Or anywhere except inside the integral, and additionally, the v in that fg part is supposed to be a v', and I don't know why that would be there. The answer should be: $$6v(0)+\int_{0}^{1}\left((x+2)v'u'+vu\right)\,dx$$
The answer is incorrect. There should be another term $\int_0^1 v \frac{du}{dx} dx$. Please verify the answer you posted.