It looks like we're given the Laplace transform of f(t),
[tex]F(p) = L_p\left\{f(t)\right\} = \dfrac6{p(2p^2+4p+10)} = \dfrac3{p(p^2+2p+5)}[/tex]
Start by splitting up F(p) into partial fractions:
[tex]\dfrac3{p(p^2+2p+5)} = \dfrac ap + \dfrac{bp+c}{p^2+2p+5} \\\\ 3 = a(p^2+2p+5) + (bp+c)p \\\\ 3 = (a+b)p^2 + (2a+c)p + 5a \\\\ \implies \begin{cases}a+b=0 \\ 2a+c=0 \\ 5a=3\end{cases} \implies a=\dfrac35,b=-\dfrac35, c=-\dfrac65[/tex]
[tex]F(p) = \dfrac3{5p} - \dfrac{3p+6}{5(p^2+2p+5)}[/tex]
Complete the square in the denominator,
[tex]p^2+2p+5 = p^2+2p+1+4 = (p+1)^2+4[/tex]
and rewrite the numerator in terms of p + 1,
[tex]3p+6 = 3(p+1) + 3[/tex]
Then splitting up the second term gives
[tex]F(p) = \dfrac3{5p} - \dfrac{3(p+1)}{5((p+1)^2+4)} - \dfrac3{5((p+1)^2+4)}[/tex]
Now take the inverse transform:
[tex]L^{-1}_t\left\{F(p)\right\} = \dfrac35 L^{-1}_t\left\{\dfrac1p\right\} - \dfrac35 L^{-1}_t\left\{\dfrac{p+1}{(p+1)^2+2^2}\right\} - \dfrac3{5\times2} L^{-1}_t\left\{\dfrac2{(p+1)^2+2^2}\right\} \\\\ L^{-1}_t\left\{F(p)\right\} = \dfrac35 - \dfrac35 e^{-t} L^{-1}_t\left\{\dfrac p{p^2+2^2}\right\} - \dfrac3{10} e^{-t} L^{-1}_t\left\{\dfrac2{p^2+2^2}\right\} \\\\ \implies \boxed{f(t) = \dfrac35 - \dfrac35 e^{-t} \cos(2t) - \dfrac3{10} e^{-t} \sin(2t)}[/tex]
