Well, Laplace transform of given function is:
(F(s)=\int_{0}^{\infty}\frac{e^{-st}}{(t+3)^2}dt)
Through partial integration with:
(u=e^{-s t},dv=\frac{1}{(t+3)^2}dt)
this gives:
(F(s)=\left.-\frac{e^{-s t}}{t+3}\right|_{0}^{\infty}-s\int_{0}^{\infty}\frac{e^{-s t}}{t+3}dt=\frac{1}{3}-s\int_{0}^{\infty}\frac{e^{-s t}}{t+3}dt)
Now through variable change:
(\mu=t+3)
this further gives:
(F(s)=\frac{1}{3}-s e^{3s}\int_{3}^{\infty}\frac{e^{-s\mu}}{\mu}d\mu)
and finally through another variable change:
(\nu=s\mu)
this gives:
(F(s)=\frac{1}{3}-s e^{3s}\int_{3s}^{\infty}\frac{e^{-\nu}}{\nu}d\nu)
which is result returned by Alpha.
