# find the Laplace transform of f (t). f(t)=\frac{e^{3t}-1}{1}

find the Laplace transform of f (t).
$f\left(t\right)=\frac{{e}^{3t}-1}{1}$
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Given
The given function is $f\left(t\right)=\frac{{e}^{3t}-1}{1}$
formula used
$L\left\{\frac{f\left(t\right)}{t}\right\}={\int }_{s}^{\mathrm{\infty }}F\left(\omega \right)d\omega$
by using the above formula
$L\left\{\frac{{e}^{3t}-1}{t}\right\}={\int }_{s}^{\mathrm{\infty }}L\left\{{e}^{3t}-1\right\}d\omega$
$={\int }_{s}^{\mathrm{\infty }}\left\{\frac{1}{\omega -3}-\frac{1}{\omega }\right\}d\omega$
$={\left[\mathrm{ln}\left(\omega -3\right)-\mathrm{ln}\left(\omega \right)\right]}_{s}^{\mathrm{\infty }}$
$={\left[\mathrm{ln}\left(\frac{\omega -3}{\omega }\right)\right]}_{s}^{\mathrm{\infty }}$
on further simplification,
$L\left\{\frac{{e}^{3t}-1}{t}\right\}=0-\mathrm{ln}\left(\frac{s-3}{s}\right)$
$=-\left[\mathrm{ln}\left(s-3\right)-\mathrm{ln}\left(s\right)\right]$
$=-\mathrm{ln}\left(s-3\right)+\mathrm{ln}\left(s\right)$
$=\mathrm{ln}\left(s\right)-\mathrm{ln}\left(s-3\right)$