This isn't an answer but a lengthy comment.

The proofs for the $H$-modul ismorphism $H \cong H^\ast$ I know of use (some variant of) the $H$-module isomorphism $I_L(H^\ast) \otimes_k H \cong H^\ast$ where
$$I_L(H^\ast) = \lbrace g \in H^\ast \mid \forall f\in H^\ast: f\ast g = f(1) \cdot g \rbrace$$ is the space of left integrals of $H^\ast$ ($f \ast g$ is the convolution). A dimension argument shows that $I_L(H^\ast)$ is one-dimensional. If $\lambda$ is a non-zero left integral then $I_L(H^\ast) = k\cdot \lambda \cong k$ yields $H \cong H^\ast$ as left $H$-modules. This is just the isomorphism described by darij.

The problem is that the proof doesn't yield a formula for such a $\lambda$. Choose a $k$-basis $\lbrace e_1=1,e_2,...,e_n\rbrace$ of $H$ and suppose $\Delta(e_k) = \sum_{i,j}d_{ij}^{(k)}\cdot e_i \otimes e_j$. Set $D^{(k)} = (d_{ij}^{(k)}-\delta_{i,1}\cdot \delta_{kj})_{i,j=1,\dots,n}$. Expressing $\lambda$ as linear combination of the dual basis then leads to the system of linear equations
$D^{(k)} x = 0$ $(k=1,...,n)$. As pointed out by Mariano it has a unique solution (up to scalars). Again, this is no closed formula, but it can be used in practise to compute integrals.

express, so it is hard to answer your question—I do doubt there is a universal expression for the isomorphism in the category generated by the structure maps of the Hopf algebra, though. – Mariano Suárez-Alvarez♦ Jan 23 '12 at 21:11