If $(M_t)$ is a Martingale, why considering $\int_0^t X_s^2d\left<M\right>_s$ instead of $\int_0^t X_s^2ds$?

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Let $(M_t)$ a martingale. In the book of Karatzas and Shreve (Brownian motion and stochastic calculs), they often consider the integral $$I_t=\int_0^t X_s^2 d \left<M\right>_s,$$ where $(\left<M\right>_s)$ is s.t. $(M_s^2-\left<M\right>_s)$ is a Martingale. I can imagine that my question can be not so precise, but in what $I_s$ is interesting ? Why considering $I_t$ rather than $\int_0^t X_s^2d s$ is interesting ?


Well, my question also work for : why considering $$\int_0^t X_s d \left<M\right>_s\quad \text{instead of }\int_0^t X_s d s \ \ ?$$