= Solution
The same <integration by parts> with $p$ replaced by $p+c$ gives
$$
\boxed{\overline{te^{-ct}}(p)=\frac1{(p+c)^2}\qquad(\operatorname{Re}(p+c)>0).}
$$
For the remaining unnumbered requests, the <derivative> and <convolution> formulas, under the usual convergence hypotheses, are
$$
\overline{f'}(p)=p\bar f(p)-f(0),\qquad\overline{f*g}(p)=\bar f(p)\bar g(p),\qquad(f*g)(t)=\int_0^tf(t-s)g(s)\,ds.
$$
Applying them to the general equation yields $p\bar x-a+\bar k\bar x=\bar y$ and therefore
$$
\boxed{\bar x(p)=\frac{\bar y(p)+a}{p+\bar k(p)}.}
$$
For the particular data, $\bar y=1/p$, $\bar k=1/(p+2)$ and $a=0$, so
$$
\bar x(p)=\frac{p+2}{p(p+1)^2}=\frac2p-\frac2{p+1}-\frac1{(p+1)^2}.
$$
The four transforms above give
$$
\boxed{x(t)=2-(2+t)e^{-t},\qquad\lim_{t\to\infty}x(t)=2.}
$$
As a direct check, $x(0)=0$, $x'=(1+t)e^{-t}$, and
$$
\int_0^te^{-2(t-s)}x(s)\,ds=1-(1+t)e^{-t}.
$$
Adding the last two expressions gives one, verifying the original equation as well as the required limit.
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