Factoring the denominator gives
[tex]s^3-2s^2+7s-14=s^2(s-2)+7(s-2)=(s-2)(s^2+7)[/tex]
(Keep in mind that [tex]7=(\sqrt7)^2[/tex].)
Split up the given transform into partial fractions:
[tex]\dfrac{3s+4}{(s-2)(s^2+7)}=\dfrac{c_1}{s-2}+\dfrac{c_2s+c_3}{s^2+7}[/tex]
[tex]3s+4=c_1(s^2+7)+(c_2s+c_3)(s-2)[/tex]
[tex]3s+4=(c_1+c_2)s^2+(-2c_2+c_3)s+7c_1-2c_3[/tex]
[tex]\implies\begin{cases}c_1+c_2=0\\-2c_2+c_3=3\\7c_1-2c_3=4\end{cases}\implies c_1=\dfrac{10}{11},c_2=-\dfrac{10}{11},c_3=\dfrac{13}{11}[/tex]
[tex]\implies\dfrac{3s+4}{(s-2)(s^2+7)}=\dfrac1{11}\left(\dfrac{10}{s-2}-\dfrac{10s-13}{s^2+7}\right)[/tex]
Recalling the frequency shift property,
[tex]F(s-a)=L(e^{at}f(t))[/tex]
we get
[tex]F(s-2)=\dfrac{10}{s-2}\implies f(t)=10e^{-2t}L^{-1}\left(\dfrac1s\right)[/tex]
[tex]\implies f(t)=10e^{-2t}[/tex]
The remaining inverse transforms reduce to sines and cosines:
[tex]F(s)=\dfrac{10s-13}{s^2+7}=\dfrac{10s}{s^2+7}-\dfrac{13}{\sqrt7}\dfrac{\sqrt7}{s^2+7}[/tex]
[tex]\implies f(t)=10\cos(\sqrt7\,t)-\dfrac{13}{\sqrt7}\sin(\sqrt7\,t)[/tex]
So we end up with
[tex]L^{-1}\left(\dfrac{3s+4}{(s-2)(s^2+7)}\right)=\boxed{\dfrac1{11}\left(10e^{-2t}-10\cos(\sqrt7\,t)+\dfrac{13}{\sqrt7}\sin(\sqrt7\,t)\right)\right)}[/tex]
