If x(t) is a solution of$(t + 1)dx=(2x+{(t + 1)}^4 ) dt, x\left(0\right)=2$ then $\underset{t\to 1}{lim} x\left(t\right)$ is

$\frac{dx}{dt}-\frac{2}{t+1}x=(t+1{)}^3$ The I. F. is $e^{-2\int \frac{dt}{t+1}}=\frac{1}{(t+1{)}^2}$

The equation reduces to  $\frac{d}{dt}\left(x\frac{1}{(t+1{)}^2}\right)=(t+1)$

$\Rightarrow \frac{x}{(t+1{)}^2}=\frac{(t+1{)}^2}{2}+C.$ Since $x\left(0\right)=h2$

so , $C=\frac{3}{2}$. Thus $\frac{x}{(t+1{)}^2}=\frac{(t+1{)}^2}{2}+\frac{3}{2}$ 

$\Rightarrow x=\frac{(t+1{)}^4}{2}+\frac{3}{2}(t+1{)}^2$.  $\underset{t\to 1}{lim} x\left(t\right)=8+6=14$