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Find $\mathrm{F}(\mathrm{t})$ if its Laplace transform is given by $f(s)=\frac{s-1}{s^{2}+2 s+10}+\frac{1}{\sqrt{s-1}}$

$F(t)=\mathcal{L}^{-1}\{f(s)\}$

$\mathcal {L}^{-1}\left\{\frac{s-1}{s^{2}+2 s+10}+\frac{1}{\sqrt{s-1}}\right\}$

$\mathcal {L}^{-1}\left\{\frac{s-1}{s^{2}+2 s+10}\right\}+\mathcal {L}^{-1}\left\{\frac{1}{\sqrt{s-1}}\right\}$

$\mathcal {L}^{-1}\left\{\frac{s-1}{s^{2}+2 s+1-1+10}\right\}+\mathcal {L}^{-1}\left\{\frac{1}{\sqrt{s-1}}\right\}$

$\mathcal{L}^{-1}\left\{\frac{s-1}{(s+1)^{2}+9}\right\}+\mathcal{L}^{-1}\left\{\frac{1}{\sqrt{s-1}}\right\}$

$S \longrightarrow S-1 \quad \quad S \longrightarrow S+1$

$e^{-t} \mathcal{L}^{-1}\left\{\frac{s-2}{s^{2}+9}\right\}+e^{t} \mathcal{L}^{-1}\left\{\frac{1}{\sqrt s}\right\}$

$\left.e^{-t}[\cos (3 t)-\frac{2}{3} \sin (3 t)\right]+e^{t} \frac{1}{\sqrt{\pi}} \cdot \frac{1}{\sqrt{t}}$

$=F(t)$

Find $F(t)$ if $L\{F(t)\}=f(s),$ and $\frac{d f}{d s}=\frac{-1}{s^{2}+4 s+13}$

$\frac{d f}{d s}=\mathcal {L}\{-t F(t)\}$

$\mathcal{L}\{-t F(t)\}=\frac{-1}{s^{2}+4 s+13}$

$\mathcal {L}^{-1} \mathcal{L}\{ F(t)\}=\mathcal {L}^{-1} \frac{1}{s^{2}+4 s+13}$

$t \cdot F(t)=\mathcal{L}^{-1}\left\{\frac{1}{s^{2}+4 s+13}\right\}$

$=\mathcal {L}^{-t}\left\{\frac{1}{s^{2}+4 s+4-4+13}\right\}$

$=\mathcal{L}^{-1}\left\{\frac{1}{s^{2}+4 s+4+9}\right\}=\mathcal {L}^{-1}\left\{\frac{1}{(s+2)^{2}+9}\right\}$

$S \rightarrow S-2$

$=e^{-2 t} \mathcal{L}^{-1}\left\{\frac{1}{s^{2}+9}\right\}=e^{-2 t} \cdot \frac{1}{3} \cdot \sin (3 t)$

$F(t)=\frac{e^{-2 t} \sin (3 t)}{3 t}$

Evaluate $L^{-1}\left\{\ln \left(1-\frac{2}{s}\right)\right\}=F(t)$

$\mathcal{L}^{-1}\{f(s)\}=F(t)$

$f(s)=\ln \left(1-\frac{2}{s}\right)=\ln \left(\frac{s-2}{s}\right)=\ln (s-2)-\ln (s)$

$\ln \left(\frac{a}{b}\right)=\ln a-\ln b$

$f^{\prime}(s)=\frac{1}{s-2}-\frac{1}{s}$

$L[-t f(t)]=\frac{1}{s-2}-\frac{1}{s}$

$\mathcal {L}^{-1} L[-t F(t)]=\mathcal {L}^{-1}\left\{\frac{1}{s-2}-\frac{1}{s}\right\}$

$-t F(t)=\mathcal {L}^{-1}\left\{\frac{1}{s-2}-\frac{1}{s}\right\}=e^{2 t}-1$

$F(t)=\frac{e^{2 t}-1}{-t}$