Understanding Itô's lemma

I am new to stochastic calculus and I am learning it during off-hours, so I am full of doubts that maybe someone more expert may dispel.

Let's say I get the usual geometric Brownian process, where $$X_0$$ is known:

$$\begin{equation}\label{1}\text dX_t=\mu X_t \text dt+\sigma X_t \text dW_t.\end{equation}$$

I apply Itô's lemma, and if $$f(x)=\ln(x)$$, I get:

$$\text df(x)=\left(\mu-\frac{\sigma^2}2\right)\text dt+\sigma \text dW_t.$$

Up to here everything is clear to me.

In the textbook I am following, they state that integrating the last equation one finds $$X_t$$:

$$X_t=X_0 \exp\left[\left(\mu-\frac{\sigma^2}2\right)dt+\sigma W_t\right].$$

If I just take the first equation, and rewrite it as $$\text d\ln(X_t)=\mu\text dt+\sigma \text dW_t,$$ and integrate, of course I miss an $$\exp\left[-\frac{\sigma^2}2t\right]$$ term. This lets me think that Itô's lemma is just the correct way to compute all the first order terms needed to perform the integration? Is this the correct way to see this?

Many thanks.

• I don't know what do you mean by "first order terms", but Ito lemma is exactly about second-order terms: you write the Taylor formula and use that $dW_t\cdot dW_t = dt$. (This must be written better in your textbook.) – zhoraster Oct 26 '18 at 10:09
• thanks a lot @zhoraster, I guess that most likely the textbook is correct and I missed something. I was thinking that this represents a correction of the first order term in time via a second order in the stochastic variable, so this is why I wrote "first order terms". – marco Oct 28 '18 at 0:11