Question on calculating hypercohomology

I want to compute the algebraic de Rham cohomology of $ \mathbb{C}^* $, and I’m confused. I don’t have much background in this, so I was hoping a very concrete example would clear up a lot of this confusion. So far:

We have this cochain of $ \mathbb{C}[x,x^{-1}]$-modules:

$0 \longrightarrow \mathbb{C}\longrightarrow \mathbb{C}[x,x^{-1}] \longrightarrow {\Omega}_{\mathbb{C}[x,x^{-1}]/\mathbb{C}}^1 \longrightarrow 0$

where ${\Omega}_{\mathbb{C}[x,x^{-1}]/\mathbb{C}}^1 = \mathbb{C}[x,x^{-1}] dx $. (as i understand)

As I understand, to obtain the de Rham cohomology we want a Cartan–Eilenberg resolution for this cochain, and then we want a left exact functor $F: \mathcal{A} \rightarrow \mathcal{B}$, where $\mathcal{A}$ is our module category.

Then we want to compute the total cohomology of $F(I)$, where $I$ is our resolution.

My issue seems to be actually doing this stuff, do I need to find $I,F$ explicitly? How would I go about doing this?

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For a smooth affine scheme $X$ of finite type over $\mathbb C$ the de Rham cohomology $H^*_{dR}(X)$ is just the cohomology of the complex of global differential forms: $$ H^*_{dR}(X)=H^*(\Gamma(X,\Omega^*_X)) $$
In your case you have to compute the cohomology of the complex $$0 \longrightarrow \mathbb{C}[x,x^{-1}] \stackrel {d}{\longrightarrow} \mathbb{C}[x,x^{-1}]\cdot dx \longrightarrow 0$$ and you immediately get: $$H^0_{dR}(X)=\mathbb C,\quad H^1_{dR}(X)=\mathbb C \cdot \frac {dx}{x}\cong \mathbb C, \quad H^i_{dR}(X)=0 \quad \text {for} \quad i\geq 2 $$ This is in line with the fact that algebraic cohomology coincides with the classical topology of the underlying complex holomorphic manifold $X^h$: $$H^*_{dR}(X)=H^*(X^h,\mathbb C)$$

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