Is a Wiener-Squared a Chi-Squared?

Question

If $W_0=0$, then $W_t\sim N(0,t)$, so $W_1^2\sim\chi_1^2$. But Ito's lemma for $f(x)=x^2$ states $\mathrm{d}\left(W_t^2\right)=\mathrm{d}t+2W_t\mathrm{d}W_t$, so $W_t^2=t+2\int_0^t{W_s\mathrm{d}W_s}$ and hence $\int_0^1{W_t\mathrm{d}W_t}=\frac{\chi^2-1}{2}$, which seems incorrect. Where am I making a mistake?