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Inverse Laplace transforms 1/(s²(s²+1))and

Short answer

Partial fractions: 1/(s²(s²+1))=1/s²−1/(s²+1), so L⁻¹ = t − sin t. For log(1+1/s²), use derivative property: L⁻¹ = 2(1−cos t)/t.

Find the inverse Laplace transform of a) 1/(s²(s² + 1)) and b) log(1 + 1/s²).

a) L⁻¹{1/(s²(s² + 1))}

Partial fractions: 1/(s²(s² + 1)) = 1/s² − 1/(s² + 1).

L⁻¹ = t − sin t

b) L⁻¹{log(1 + 1/s²)}

Let F(s) = ln((s² + 1)/s²) = ln(s² + 1) − 2 ln s. Using L{t f(t)} = −F'(s):

F'(s) = 2s/(s² + 1) − 2/s
L{t f(t)} = 2/s − 2s/(s² + 1)
t f(t) = 2 − 2 cos t
f(t) = 2(1 − cos t)/t

Answers: a) t − sin t; b) 2(1 − cos t)/t.

✓ Answer checked by Claude Opus 5.5 (high effort)

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