Given that L[f(t)]=s+2s2+1,L[f(t)]=s2+1(s+3)(s+2),\mathcal{L}[f(t)]=\frac{s+2}{s^2+1}, \mathcal{L}[f(t)]=\frac{s^2+1}{(s+3)(s+2)},L[f(t)]=s2+1s+2,L[f(t)]=(s+3)(s+2)s2+1, h(t)=∫01f(τ)g(t−τ)dτ,L[h(t)]h(t)=\int_0^1 f(\tau) g(t-\tau) d \tau, \mathcal{L}[h(t)]h(t)=∫01f(τ)g(t−τ)dτ,L[h(t)] isChoose one optionAs2+1s+3\frac{s^2+1}{s+3}s+3s2+1B1s+3\frac{1}{s+3}s+31Cs2+1(s+3)(s+2)+s+2s2+1\frac{s^2+1}{(s+3)(s+2)}+\frac{s+2}{s^2+1}(s+3)(s+2)s2+1+s2+1s+2DNone of theseSubmit50:50Try againView answerShowHideAnswer: BWorked explanation not available yet.Free practice · Your answers on this page are not saved.