Encontre a transformada inversa: a) L−1{ 1s2 − 1s + 1s−2} b) L−1{ 14s+1} c) L−1{ s−1s2+2} d) L−1{ s+1(s2−4s)(s+5)} e) L−1{ 6s+3(s2+1)(s2+4)}

a) L−1{ 1s2 − 1s + 1s−2} = t - 1 + u(t), onde u(t) é a função degrau unitário. b) L−1{ 14s+1} = 14e^(-t) c) L−1{ s−1s2+2} = cos(sqrt(2)t) d) L−1{ s+1(s2−4s)(s+5)} = (1/2)e^(4t) - (1/2)e^(-5t) + (1/2)delta(t) e) L−1{ 6s+3(s2+1)(s2+4)} = 3cos(2t) + 6sen(t) + 3cos(4t)