# Using the input integral principle below ccL[int_0^t f(u)du](s)=1/s ccL[f(t)](s), s>c Find ccL^(-1)[(1)/(s(s^2+1))](t) without using partial fractions.

Using the input integral principle below

Find without using partial fractions.
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Katelyn Chapman
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$\mathcal{L}\left[{\int }_{0}^{t}f\left(u\right)du\right]\left(s\right)=\frac{1}{s}\mathcal{L}\left[f\left(t\right)\right]\left(s\right)$
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GepGreeloCesyjk
Hint: The implied way is
$\mathcal{L}\left[{\int }_{0}^{t}f\left(u\right)du\right]\left(s\right)=\frac{1}{s}\mathcal{L}\left[f\left(t\right)\right]\left(s\right)=\frac{1}{s}\frac{1}{{s}^{2}+1}\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}\mathcal{L}\left[f\left(t\right)\right]\left(s\right)=\frac{1}{{s}^{2}+1}.$