$(B_t)_{t \geq 0}$ is a standard Brownian motion, first of all I know that they are equal in distribution but as I was solving the Geometric Brownian motion SDE I got a solution of the form :
$$X_t = X_0 \exp((r - \frac{\sigma^2}{2})t + \sigma\int_{0}^{t} dB_s)$$
but the solution that is written in the lecture notes I'm reading from is of the form :
$$X_t = X_0 \exp((r - \frac{\sigma^2}{2})t + \sigma B_t)$$
Am I wrong and should review my work or are the two forms the same thing ?