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# Find the Laplace of {$\frac{1}{t+1}$}

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Definition of Laplace Transform:

{$$F(s)=\int _0^\infty f(t)e^{-st}\, dt\$$}

To find the Laplace transform of {$\displaystyle f(t) = \frac{1}{t+1}$}

{\begin{align} F(s)=\int _0^\infty \frac{1}{t+1}e^{-st}\, dt\ \cr \end{align}}

Integration by parts:

{$$\large \int uv' dx = uv - \int u'v dx \;$$}

 ' {$u$} {$\displaystyle \quad e^{-st} \quad$} {$\displaystyle \quad -se^{-st}\quad$} {$v$} {$\displaystyle \quad \ln (1+t) \quad$} {$\displaystyle \frac{1}{t+1}\quad$}

{$$\large \int _0^\infty uv' dt = \left.uv\right| _0^\infty - \int _0^\infty u'v dt \;$$}

{$$\large \int _0^\infty \frac{1}{t+1}e^{-st} dt = \left.ln(t+1) e^{-s t}\right| _0^\infty - \int _0^\infty \ln(t+1)\left(-\frac{{{e}^{-s t}}}{s}\right) dt \;$$}

{$$\large \int _0^\infty \frac{1}{t+1}e^{-st} dt = \lim _{R \rightarrow \infty} \left.ln(t+1) e^{-s t}\right| _0^R - \int _0^\infty \ln(t+1)\left(-\frac{{{e}^{-s t}}}{s}\right) dt \;$$}

{$$\large \int _0^\infty \frac{1}{t+1}e^{-st} dt = \left.\frac{1}{t+1}\left(-\frac{{{e}^{-s t}}}{s}\right)\right| _0^\infty - \int _0^\infty \frac{e^{-st}}{{s\left( t+1\right) }^{2}} dt \;$$}

{$$\displaystyle \frac{{{e}^{-s t}}}{s\, {{t}^{2}}+2 s t+s}$$}

{$$\displaystyle -\frac{1}{{{e}^{s t}} s t+{{e}^{s t}} s}$$}

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