If $$x = x(t)$$ is the solution of the differential equation $$(t+1)dx = (2x + (t+1)^4)dt$$, $$x(0) = 2$$, then $$x(1)$$ equals:

The given differential equation is $$(t+1) dx = (2x + (t+1)^4) dt$$ with initial condition $$x(0) = 2$$.

Rewrite the equation as: $$(t+1) dx - 2x dt = (t+1)^4 dt$$

Divide both sides by $$(t+1)^3$$: $$\frac{(t+1) dx - 2x dt}{(t+1)^3} = (t+1) dt$$

The left side is the differential of $$\frac{x}{(t+1)^2}$$ because: $$d\left( \frac{x}{(t+1)^2} \right) = \frac{(t+1) dx - 2x dt}{(t+1)^3}$$

Thus, the equation becomes: $$d\left( \frac{x}{(t+1)^2} \right) = (t+1) dt$$

Integrate both sides: $$\int d\left( \frac{x}{(t+1)^2} \right) = \int (t+1) dt$$ $$\frac{x}{(t+1)^2} = \frac{(t+1)^2}{2} + C$$

Solve for $$x$$: $$x = (t+1)^2 \left( \frac{(t+1)^2}{2} + C \right) = \frac{(t+1)^4}{2} + C (t+1)^2$$

Apply the initial condition $$x(0) = 2$$: Substitute $$t = 0$$ and $$x = 2$$: $$2 = \frac{(0+1)^4}{2} + C (0+1)^2 = \frac{1}{2} + C$$ $$C = 2 - \frac{1}{2} = \frac{3}{2}$$

Thus, the solution is: $$x(t) = \frac{(t+1)^4}{2} + \frac{3}{2} (t+1)^2$$

Evaluate $$x(1)$$: Substitute $$t = 1$$: $$x(1) = \frac{(1+1)^4}{2} + \frac{3}{2} (1+1)^2 = \frac{16}{2} + \frac{3}{2} \times 4 = 8 + 6 = 14$$

Therefore, $$x(1) = 14$$.