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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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