By the convolution theorem,
[tex]f(t)=(g*h)(t)=\displaystyle\int_{\tau=0}^{\tau=t}g(\tau)h(t-\tau)\,\mathrm d\tau\implies F(s)=G(s)H(s)[/tex]
where [tex]F(s),G(s),H(s)[/tex] are the respective Laplace transforms of [tex]f(t),g(t),h(t)[/tex].
We have
[tex]F(s)=\dfrac1{s^5(s^2+1)}=\dfrac1{4!}\dfrac{4!}{s^5}\times\dfrac1{s^2+1}[/tex]
If we let [tex]G(s)=\dfrac{4!}{s^5}[/tex] and [tex]H(s)=\dfrac1{s^2+1}[/tex], then clearly [tex]g(t)=t^4[/tex] and [tex]h(t)=\sin t[/tex], so by the convolution theorem,
[tex]f(t)=\dfrac1{4!}t^4*\sin t=\displaystyle\frac1{24}\int_{\tau=0}^{\tau=t}\tau^4\sin(t-\tau)\,\mathrm d\tau=\frac{t^4}{24}-\frac{t^2}2+1-\cos t[/tex]
