If $x=x(t)$ is the solution of the differential equation $(t+1) \mathrm{d} x=\left(2 x+(t+1)^4\right) \mathrm{dt}, x(0)=2$, then, $x(1)$ equals _________.

$\begin{aligned} & (\mathrm{t}+1) \mathrm{dx}=\left(2 \mathrm{x}+(\mathrm{t}+1)^4\right) \mathrm{dt} \\\\ & \frac{\mathrm{dx}}{\mathrm{dt}}=\frac{2 \mathrm{x}+(\mathrm{t}+1)^4}{\mathrm{t}+1} \\\\ & \frac{\mathrm{dx}}{\mathrm{dt}}-\frac{2 \mathrm{x}}{\mathrm{t}+1}=(\mathrm{t}+1)^3\end{aligned}$ <br/><br/>$I.F=e^{-\int \frac{2}{t+1} d t}=e^{-2 \ln (t+1)}=\frac{1}{(t+1)^2}$ <br/><br/>$\begin{aligned} & \frac{x}{(t+1)^2}=\int \frac{1}{(t+1)^2}(t+1)^3 d t+C\end{aligned}$ <br/><br/>$\frac{x}{(t+1)^2}=\frac{t^2}{2}+t+C$ <br/><br/>Now $x(0)=2$ <br/><br/>$$ \begin{aligned} & \Rightarrow C=2 \\\\ & \therefore x=\left(\frac{t^2}{2}+t+2\right)(t+1)^2 \\\\ & x(1)=\left(\frac{1}{2}+1+2\right)(1+1)^2 \\\\ & =\frac{7}{2} \times 4=14 \end{aligned} $$