Complete the square in the denominator to reveal a sum of squares:
[tex]s^2+4s+29=(s+2)^2+25=(s+2)^2+5^2[/tex]
Recall the Laplace transforms of sine and cosine,
[tex]L(\sin(at))=\dfrac a{s^2+a^2}[/tex]
[tex]L(\cos(at))=\dfrac s{s^2+a^2}[/tex]
as well as the frequency shift property,
[tex]L(e^{at}f(t))=F(s-a)[/tex]
where [tex]L(f(t))=F(s)[/tex] is the Laplace transform of [tex]f(t)[/tex].
Rewrite the given transform as
[tex]\dfrac{3s+4}{s^2+4s+29}=\dfrac{3(s+2)}{(s+2)^2+5^2}-\dfrac25\dfrac5{(s+2)^2+5^2}[/tex]
The inverse transforms then follows:
[tex]F(s+2)=\dfrac{3(s+2)}{(s+2)^2+5^2}\implies f(t)=3e^{-2t}L^{-1}\left(\dfrac s{s^2+5^2}\right)[/tex]
[tex]\implies f(t)=3e^{-2t}\cos(5t)[/tex]
[tex]F(s+2)=-\dfrac25\dfrac5{(s+2)^2+5^2}\implies f(t)=-\dfrac25e^{-2t}L^{-1}\left(\dfrac5{(s+2)^2+5^2}\right)[/tex]
[tex]\implies f(t)=-\dfrac25e^{-2t}\sin(5t)[/tex]
So we end up with (after some regrouping)
[tex]L^{-1}\left(\dfrac{3s+4}{s^2+4s+29}\right)=\boxed{\dfrac{e^{-2t}}5(15\cos(5t)-2\sin(5t))}[/tex]
