Show that geometric brownian motion is solution to the stochastic differential equation

Question

I want to show that $X_t = X_0 e^{\sigma B_t + \left (\mu - 1/2\sigma^2 \right)t}$ together with $(B_t)_{t\geq 0}$ and the filtration $(\mathcal{F}_t)_{t\geq0}$ solves the SDE $$ X_t = X_0 + \int_0^t \mu X_r\,dr + \int_0^t \sigma X_r \,dB_r. $$ My idea is to use Itô's formula with the twice continuously differentiable function $F(t,X_t) = \ln(X_t)$, however, how is this done in such a case? Itô's formula is given by: $$ F(X_t) - F(X_0) = \int_0^t F'(X_r)\,dX_r + \frac{1}{2}\int_0^t F''(X_r)\,d[X]_r $$ Firstly, the quadratic variation of $X$: \begin{aligned} \left [ X\right]_t = [\mu \cdot t]_t + [\sigma \cdot B]_t = \int_0^t \mu^2 \,ds + \int_0^t \sigma^2 \, [B]_s = \int_0^t \sigma^2 \,ds \end{aligned} Which gives us $$ \ln(X_t) - \ln(X_0) = \int_0^t \frac{1}{X_r} \,dX_r - \frac{1}{2} \int_0^t \frac{1}{X_r^2} \sigma^2 \,dr $$ Tips for proceeding?, I am quite new to Itô's formula and stochastic calculus.