We need the equation $3\sin x + 4\cos x = k+1$ to have at least one real solution in $x$.

The expression $3\sin x + 4\cos x$ can be written in the form $R\sin(x+\phi),$ where $R=\sqrt{3^2+4^2}=5.$ Hence, $3\sin x + 4\cos x$ has range $[-5,5].$

Therefore, for the equation $3\sin x + 4\cos x = k+1$ to have a solution, the right-hand side must satisfy $-5 \le k+1 \le 5.$

Subtracting $1$ throughout, $-6 \le k \le 4.$

Now count the integral values of $k$ in this interval: $-6,-5,-4,-3,-2,-1,0,1,2,3,4$ Total number of integers $= 4-(-6)+1 = 11.$

Hence, the required number of integral values of $k$ is $\boxed{11}.$