Integration by parts to find the adjoint operator On the interval $(0,1)$ consider the differential operator $Lu=u''''+u'$ with boundary conditions
$u(0)+u'(1)=u(1)+u'(0)=0$
$2u(0)+u''(1)=2u(1)+u''(0)=0$  
$(1)$
I want to find the adjoint operator and the adjoint boundary conditions.
Or more precisely, I want to find a differential operator $L*$ and four vectors $C_{j}\in\mathbb{R}^{8}$ so that for any $v\in C^{4}([0,1],\mathbb{R})$, the following are equivalent
i) $\displaystyle \int_{0}^{1} v(x)(Lu)(x)dx=\int_{0}^{1}(L^{\ast}v)(x)dx$ $\forall u\in C^{4}([0,1],\mathbb{R})$ that satisfy $(1)$.
ii) $\forall 1 \le j \le 4$ we have $C_{j}\cdot(v(0),v'(0), v''(0), v'''(0), v(1), v'(1), v''(1), v'''(1))=0$.
So using the fact that $(Lu,v)=(u,L^{\ast}v)$, we have that
$\displaystyle (Lu,v)=\int_{0}^{1}(Lu)(x)v(x)dx=\int_{0}^{1}v(x)(u''''(x)+u'(x))dx$
Then using integration by parts, we eventually end up with
$\displaystyle (Lu,v)=v(x)u(x)+v(x)u'''(x)-v'(x)u''(x)+v''(x)u'(x)-v'''(x)u(x)+\int_{0}^{1}v''''(x)u(x)-v'(x)u(x)dx$
What do I do from here?
 A: (Too long for a comment so I'll put this here) You last comment was heading in the right direction - with the first term $\int_0^1 v(x)u''''(x)\mathrm{d}x$, you integrate by parts once to get 
\begin{equation}[v(x)u'''(x)]_0^1 - \int_0^1 v'(x)u'''(x)\mathrm{d}x
\end{equation} 
as you said (although you're missing the boundary terms that would come from integrating the second term by parts, I think). We don't know anything about $u'''$ or $v$ at the boundaries (yet), so we'll leave the first part alone and integrate by parts again to get \begin{equation}
[v(x)u'''(x)]_0^1 - [v'(x)u''(x)]_0^1+\int_0^1 v''(x)u''(x)\mathrm{d}x.
\end{equation} Continuing this to the obvious conclusion, we end up with 
\begin{equation}
[v(x)u'''(x)]_0^1 - [v'(x)u''(x)]_0^1 + [v''(x)u'(x)]_0^1 - [v'''(x)u(x)]_0^1 +\int_0^1 v''''(x)u(x)\mathrm{d}x
\end{equation}
From the other term, we get 
\begin{equation}
\int_0^1 v(x)u'(x)\mathrm{d}x = [v(x)u(x)]_0^1 - \int_0^1 v'(x)u(x)\mathrm{d}x.
\end{equation}
Now putting these together, we get 
\begin{equation}
\int_0^1 v(x)(u''''(x) +u'(x))\mathrm{d}x = [v(x)u'''(x)]_0^1 - [v'(x)u''(x)]_0^1 + [v''(x)u'(x)]_0^1 - [v'''(x)u(x)]_0^1 + [v(x)u(x)]_0^1 +\int_0^1 (v''''(x)-v'(x))u(x)\mathrm{d}x,
\end{equation}
so your adjoint operator is clearly $L^*v=v''''-v'$, and you can plug in the boundary conditions on $u$ and solve to get the correct boundary conditions on $v$. I don't have a pen and paper handy so I can't do that part myself at the moment but it shouldn't be too difficult from there. Hope that helps. 
