$$\mathcal{L}[{\int}_{0}^{t}f(u)du](s)=\frac{1}{s}\mathcal{L}[f(t)](s),\text{}sc$$

Find ${\mathcal{L}}^{-1}\left[\frac{1}{s({s}^{2}+1)}\right](t)\text{}$ without using partial fractions.

easternerjx
2022-09-25
Answered

Using the input integral principle below

$$\mathcal{L}[{\int}_{0}^{t}f(u)du](s)=\frac{1}{s}\mathcal{L}[f(t)](s),\text{}sc$$

Find ${\mathcal{L}}^{-1}\left[\frac{1}{s({s}^{2}+1)}\right](t)\text{}$ without using partial fractions.

$$\mathcal{L}[{\int}_{0}^{t}f(u)du](s)=\frac{1}{s}\mathcal{L}[f(t)](s),\text{}sc$$

Find ${\mathcal{L}}^{-1}\left[\frac{1}{s({s}^{2}+1)}\right](t)\text{}$ without using partial fractions.

You can still ask an expert for help

Katelyn Chapman

Answered 2022-09-26
Author has **13** answers

Compare these

$$\mathcal{L}[{\int}_{0}^{t}f(u)du](s)=\frac{1}{s}\mathcal{L}[f(t)](s)$$

$$\mathcal{L}\left[{\int}_{0}^{t}{\mathrm{sin}u}\text{}du\right](s)=\frac{1}{s}{\frac{1}{({s}^{2}+1)}}$$

$$\mathcal{L}[{\int}_{0}^{t}f(u)du](s)=\frac{1}{s}\mathcal{L}[f(t)](s)$$

$$\mathcal{L}\left[{\int}_{0}^{t}{\mathrm{sin}u}\text{}du\right](s)=\frac{1}{s}{\frac{1}{({s}^{2}+1)}}$$

GepGreeloCesyjk

Answered 2022-09-27
Author has **1** answers

Hint: The implied way is

$$\mathcal{L}[{\int}_{0}^{t}f(u)du](s)=\frac{1}{s}\mathcal{L}[f(t)](s)=\frac{1}{s}\frac{1}{{s}^{2}+1}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}\mathcal{L}[f(t)](s)=\frac{1}{{s}^{2}+1}.$$

$$\mathcal{L}[{\int}_{0}^{t}f(u)du](s)=\frac{1}{s}\mathcal{L}[f(t)](s)=\frac{1}{s}\frac{1}{{s}^{2}+1}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}\mathcal{L}[f(t)](s)=\frac{1}{{s}^{2}+1}.$$

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