Here comes from a proof in Le Gall's Book Brownian Motion and Stochastic Calculus:
Let $D$ be a continuous local martingale taking positive values. There exists a unique continuous local martingale $L$ such that \begin{equation} D_{t}=\exp\left(L_{t}-\frac{1}{2}\left\langle L,L\right\rangle_{t}\right)=\mathcal{E}(L)_{t}. \end{equation} Moreover, $L$ is given by the formula \begin{equation} L_{t}=\log D_{0}+\int_{0}^{t}\frac{1}{D_{s}}\text{d}D_{s}. \end{equation}
In the proof, we can apply Ito's formula to $\log D_{t}$, and we get \begin{equation} \log D_{t}=\log D_{0}+\int_{0}^{t}\frac{\text{d}D_{s}}{D_{s}}-\frac{1}{2}\frac{\text{d}\left\langle D,D\right\rangle_{s}}{D_{s}^{2}}=L_{t}-\frac{1}{2}\left\langle L,L\right\rangle_{t} \end{equation} My question is, how can we directly set $L_{t}=\log D_{0}+\int_{0}^{t}\frac{\text{d}D_{s}}{D_{s}}$, and why $\frac{1}{2}\frac{\text{d}\left\langle D,D\right\rangle_{s}}{D_{s}^{2}}=\frac{1}{2}\left\langle L,L\right\rangle_{t}$ Please show me the calculation,thanks!