TITLE: de Rham cohomology of smooth affine varieties
QUESTION [9 upvotes]: Let $U$ be a smooth variety over $\mathbb{C}$. We know that there exists a smooth compactification $X$ such that $X-U$ is a normal crossings divisor $D$ and that the de Rham cohomology of $U$ can be computed using the complex of sheaves with logarithmic differentials on $X$:
$$
H_{dR}^\ast(U)=\mathbb{H}^\ast(X, \Omega^\bullet_X(\log D))
$$
In general, one needs hypercohomology here but I am wondering if, when $U$ is affine, one simply has
$$
H^\ast_{dR}(U)=H^\ast(\Gamma(\Omega^\bullet_X(\log D))
$$ I know that this is true if one uses instead the de Rham complex $\Omega^\bullet_U$ to compute cohomology and I know that $j_\ast \Omega_U^\bullet$ and $\Omega_X^\bullet(\log D)$ are quasi-isomorphic (here $j: U \hookrightarrow X$). Is this enough to conclude?

REPLY [4 votes]: The de Rham cohomology of an affine variety $U$ is simply $H^\bullet(\Gamma(U,\Omega_U^\bullet))$ where $\Omega_U^\bullet$ is the sheaf of algebraic forms. See Grothendieck (1966), « On the de Rham cohomology of algebraic varieties », Inst. Hautes Études Sci. Publ. Math., (29), p. 96.
In general, you cannot get rid of hypercohomology AND use logarithmic forms only.