If [tex]F(s)[/tex] is the Laplace transform of [tex]f(t)[/tex], then first recall the phase shift property of the transform:
[tex]\mathcal L_s\{e^{ct}f(t)\}=F(s-c)[/tex]
In this case, [tex]c=-1[/tex], and [tex]F(s)[/tex] is the transform of [tex]f(t)=9t\sin3t[/tex].
We can easily (if tediously) derive the following result:
[tex]\mathcal L_s\{t\sin(at)\}=\dfrac{2as}{(s^2+a^2)^2}[/tex]
so that
[tex]F(s)=\dfrac{9(6s)}{(s^2+9)^2}=\dfrac{54s}{(s^2+9)^2}[/tex]
and so
[tex]F(s+1)=\mathcal L_s\left\{9te^{-t}\sin3t\right\}=\dfrac{54(s+1)}{((s+1)^2+9)^2}[/tex]
[tex]F(s+1)=\dfrac{54s+54}{s^4+4s^3+24s^2+40s+100}[/tex]
