If we know the dynamics of $S$, then we can estimate the value of $S$ at a time point, $t$. Here, I have a question concerning how to solve for $S_t$ by Itô because I obtained different results by different approaches.
For a geometric Brownian motion: $$dS_t=S_t μ dt+S_t σdW_t,$$ $$\frac{dS_t}{S_t} =μ dt+σdW_t,$$ and, in fact we have, $$\frac{dS_t}{S_t} =d\ln(S_t).$$ If we make $Z=d\ln(S_t)$, then, $$dZ=\frac{\partial Z}{\partial t} dt+\frac{\partial Z}{\partial S_t} dS_t+ \frac{1}{2} \frac{\partial^2 Z}{\partial S_t^2} (dS_t)^2=(μ-\frac{1}{2} σ^2 )dt+σdW_t,$$
$$Z_t= Z_0+\left(μ- \frac{σ^2}{2} \right) \int_0^tds+σ\int_0^tdW_s,$$ $$\ln(S_t )=\ln(S_0 )+(μ-\frac{1}{2} σ^2 )dt+σW_t,$$ $$S_t=S_0 \cdot e^\left((μ- \frac{1}{2} σ^2 )dt+σW_t \right).$$
However, if I use another approach, then I get the different result. Since we have $\frac{dS_t}{S_t} =d\ln(S_t)$ then, $$d\ln(S_t )=μdt+σdW_t$$ and \begin{align} \ln(S_t)&=\ln(S_0)+μ\int_0^tds+σ\int_0^t dW_s \\ &=\ln(S_0)+μt + σW_t, \end{align} $$S_t=S_0 \cdot e^{(μt+σW_t)}$$
I think both approaches are correct. But why are the results distinct?