Step 1: Split the integral into two parts.
$\int_{-1}^{1}(x[1+\sin(\pi x)]+1)\,dx = \int_{-1}^1 x[1+\sin(\pi x)]\,dx + \int_{-1}^1 1\,dx$. The second integral equals $2$.
Step 2: Analyze $g(x) = x[1+\sin(\pi x)]$ on $(-1, 0)$.
For $x \in (-1, 0)$: $\pi x \in (-\pi, 0)$, so $\sin(\pi x) \in (-1, 0)$ (negative). Hence $1 + \sin(\pi x) \in (0, 1)$, and the greatest integer $[1+\sin(\pi x)] = 0$. So $g(x) = x \cdot 0 = 0$ on $(-1, 0)$.
Step 3: Evaluate $\int_{-1}^0 g(x)\,dx$.
Since $g(x) = 0$ on $(-1, 0)$ (the endpoints form a set of measure zero and do not affect the integral), $\int_{-1}^0 g(x)\,dx = 0$.
Step 4: Analyze $g(x) = x[1+\sin(\pi x)]$ on $(0, 1)$.
For $x \in (0, 1)$: $\pi x \in (0, \pi)$, so $\sin(\pi x) \in (0, 1]$ (positive). Hence $1 + \sin(\pi x) \in (1, 2]$, and the greatest integer $[1+\sin(\pi x)] = 1$ (since $\sin(\pi x) < 1$ for all $x \in (0,1)$ except $x=1/2$ where it equals $1$, so $1+\sin(\pi x) < 2$ except at one point). So $g(x) = x \cdot 1 = x$ on $(0, 1)$.
Step 5: Evaluate $\int_0^1 g(x)\,dx$.
$\int_0^1 x\,dx = \left[\frac{x^2}{2}\right]_0^1 = \frac{1}{2}$.
Step 6: Combine all parts.
$\int_{-1}^1 g(x)\,dx = 0 + \frac{1}{2} = \frac{1}{2}$. Total $= \frac{1}{2} + 2 = \frac{5}{2}$. \[\boxed{\dfrac{5}{2}}\]