Let $T\colon H^1_0(\mathbb{R})\to \mathbb{R}, \;T(f)=\int f'\phi \;dx$, where $\phi\in L^2$ is fixed. By Hölder, $|T(f)|\leq\|\phi\|_2\|f'\|_2\leq C \|f\|_{H^1_0}$, i.e. $T$ is continuous. Therefore, by the Riesz representation theorem there should exist $\psi\in H^1_0$, such that $T(f)=<f,\psi>_{H^1_0}=\int f\psi + f'\psi' \; dx \quad \forall f\in H^1_0$.
My problem is that I don't see how this can be possible. On the other hand, I can't prove it is not possible, so I am willing to believe it but would be thankful for any explanatory remarks/good intuition on that.
Things make more sense when you look at the Fourier transform. Letting $\widehat{\phi}$ and $\widehat{f}$ be the Fourier transforms of $\phi$ and $f$, and ignoring dimensional constants $$ \int f' \phi = \int i\xi \widehat{f} \widehat{\phi}. $$ On the other hand, $$ \int f\psi + f'\psi' = \int (1 + |\xi|^2) \widehat{f} \widehat{\psi}. $$ If we take $\widehat{\psi} = i\xi \widehat{\phi} / (1 + |\xi|^2)$, then these two expressions match. This formula tells you that this would also work if $f\in H^{-1}(\mathbb{R})$ instead of just $L^2$.
As for the intuition, this formula makes a little bit of sense if you try to find $\psi$ using integration by parts. The first step is to observe $$ \int f' \phi = -\int f\phi' $$ and this suggests that $\psi = \phi'$ might be a good first guess. Maybe this explains the multiplication by $i\xi$. Now that guess for $\psi$ fails because you can't control the $f'\psi'$ term. Maybe that explains the $(1+|\xi|^2)^{-1}$ term. It would be interesting to think more about this ...