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Inverse Laplace transforms (s+2)/(s+1)⁴

Short answer

i) (s+2)/(s+1)⁴ = 1/(s+1)³ + 1/(s+1)⁴; inverse is e^(−t)(t²/2 + t³/6). ii) Differentiate cot⁻¹(s+1) to −1/(( s+1)²+1); L{t f(t)} = −F'(s) gives t f(t)=e^(−t) sin t, so f(t)=e^(−t) sin t / t.

Find the inverse Laplace transform of: i) (s + 2)/(s + 1)⁴ ii) cot⁻¹(s + 1).

i) L⁻¹{(s + 2)/(s + 1)⁴}

Write s + 2 = (s + 1) + 1:

(s + 2)/(s + 1)⁴ = 1/(s + 1)³ + 1/(s + 1)⁴

Using L⁻¹{1/(s + 1)ⁿ} = e^(−t) t^(n−1)/(n − 1)!:

L⁻¹ = e^(−t) t²/2 + e^(−t) t³/6 = e^(−t) (t²/2 + t³/6)

ii) L⁻¹{cot⁻¹(s + 1)}

Let F(s) = cot⁻¹(s + 1). Then F'(s) = −1/((s + 1)² + 1), and with L{t f(t)} = −F'(s):

L{t f(t)} = 1/((s + 1)² + 1)  → t f(t) = e^(−t) sin t
f(t) = e^(−t) sin t / t

Answers: i) e^(−t)(t²/2 + t³/6); ii) e^(−t)(sin t)/t.

✓ Answer checked by Claude Opus 5.5 (high effort)

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