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Inverse Laplace transforms 1/(s(s+2))

Short answer

For 1/(s(s+2)), partial fractions give ½(1/s -1/(s+2)), so inverse is (1 - e^(-2t))/2. For tan^(-1)(1/s), differentiate: F'(s) = -1/(s^2+1), thus t f(t) = sin t, so f(t) = sin t / t.

Find the inverse Laplace transform of (i) 1/(s(s + 2)) (ii) tan⁻¹(1/s).

(i) L⁻¹[1/(s(s + 2))]

1/(s(s+2)) = (1/2) [ 1/s − 1/(s + 2) ]

Answer: (1 − e^(−2t))/2

(ii) L⁻¹[tan⁻¹(1/s)]

Let F(s) = tan⁻¹(1/s). Then F′(s) = 1/(1 + 1/s²) = −1/(s² + 1). Since L[t f(t)] = −F′(s) = 1/(s² + 1), we have t f(t) = sin t.

Answer: f(t) = sin t / t

✓ Answer checked by Claude Opus 5.5 (high effort)

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