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

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