Given a smooth projective curve $C$ over a field $k$ of characteristic $0$, let $D$ be a divisor. Let $\Omega_C^{\bullet}(D)$ be the logarithmic de Rham complex of $C$, with log poles along $D$.
Is it true that
$$(*) \; \;H^1_{dR}(C, \Omega^{\bullet}_C(D))= H^1(C, \mathcal{O}_C) \oplus H^0(C, \Omega^1_C(D))$$
I think it is true, for the following reasons. I'm able to show that if we take the standard dummy filtration $F$ on the logarithmic de Rham complex, then we find that
$$ F^0H^1_{dR}(C, \Omega^{\bullet}_C(D))=H^1_{dR}(C, \Omega^{\bullet}_C(D))$$ $$ F^1 H^1_{dR}(C, \Omega^{\bullet}_C(D)) = H^0(C, \Omega^1_C(D))$$ $$ F^p H^1_{dR}(C, \Omega^{\bullet}_C(D)) =0 \text{ for } p>1$$
Since $C$ is a curve then $\Omega^p_C(D)=0$ for $p>1$. Let $K^{\bullet}$ be $Tot^{\bullet}(C^{\bullet}(\Omega_C^{\bullet}(D)))$, the total complex associated to the Godemont resolution of $\Omega_C^{\bullet}(D)$. Then the associated filtration on $K{\bullet}$ is such that $F^p K^{\bullet}=0$ for $p>1$.
Then in the spectral sequence associated to this filtration we have
$$E^{p,q}_1 := R^{p+q}(\Gamma(X,-)(Gr_F^p(K^{\bullet}))) = H^q(C, \Omega^p_C(D))$$
Then note that $E^{0,1}_1 = H^1(C, \mathcal{O}_C)$ and $E^{1,0}_1 = H^0(C, \Omega^1_C(D))$. However, all the other terms of the spectral sequence are $0$, since the filtration is concentrated in degrees $0,1$. So then $E^{p,q}_1=0$ for $p>1$. Additionally, since $C$ is a curve, $E^{pq}_1=0$ for $q>1$. Now
$$ E^{pq}_2 = H\left( E^{p-1,q}_1 \xrightarrow{d} E^{p,q}_1 \xrightarrow{d} E^{p+1,q}_1 \right)$$.
Hence
$$E^{0,q}_2 = \ker \left( H^q(C, O_C) \xrightarrow{d} H^q(C, \Omega^1_C(D))\right)$$ $$E^{1,q}_2 = \dfrac{H^q(C, \Omega^1_C(D))}{ \text{im}\left(H^q(C, O_C) \xrightarrow{d} H^q(C, \Omega^1_C(D))\right)}$$ $$E^{p,q}_2 = 0 \text{ for }p >1$$.
and it is clear that $E_2=E_{\infty}$. What I think I need to show is that the spectral sequence actually degenerates at $r=1$, so that $E_1=E_{\infty}$. Then we would have $E^{p,q}_1= Gr_F^p(H^{p+q}(K^{\bullet})$, and so $E^{0,1}_1 = Gr^0_F(H^1(K^{\bullet}))$ and $E^{1,0}_1 = Gr^1_F(H^1(K^{\bullet}))$, and we'd be able to deduce $(*)$.