Problem 02 | Division by t

$f(t) = \dfrac{e^{4t} - e^{-3t}}{t}$

$f(t) = \dfrac{e^{4t}}{t} - \dfrac{e^{-3t}}{t}$

$\mathcal{L} \left\{ f(t) \right\} = \mathcal{L} \left\{ \dfrac{e^{4t}}{t} \right\} - \mathcal{L} \left\{ \dfrac{e^{-3t}}{t} \right\}$
 

Since
$\mathcal{L} (e^{4t}) = \dfrac{1}{s - 4}$   and

$\mathcal{L} (e^{-3t}) = \dfrac{1}{s + 3}$
 

Then,
$\displaystyle \mathcal{L} \left\{ f(t) \right\} = \int_s^\infty \dfrac{du}{u - 4} - \int_s^\infty \dfrac{du}{u + 3}$

$\mathcal{L} \left\{ f(t) \right\} = \Big[ \ln (u - 4) \Big]_s^\infty - \Big[ \ln (u + 3) \Big]_s^\infty$

$\mathcal{L} \left\{ f(t) \right\} = \Big[ \ln (u - 4) - \ln (u + 3) \Big]_s^\infty$

$\mathcal{L} \left\{ f(t) \right\} = \left[ \ln \dfrac{u - 4}{u + 3} \right]_s^\infty$

$\mathcal{L} \left\{ f(t) \right\} = \left[ \ln \dfrac{\dfrac{u}{u} - \dfrac{4}{u}}{\dfrac{u}{u} + \dfrac{3}{u}} \right]_s^\infty$

$\mathcal{L} \left\{ f(t) \right\} = \left[ \ln \dfrac{1 - \dfrac{4}{u}}{1 + \dfrac{3}{u}} \right]_s^\infty$

$\mathcal{L} \left\{ f(t) \right\} = \ln \dfrac{1 - \dfrac{4}{\infty}}{1 + \dfrac{3}{\infty}} - \ln \dfrac{1 - \dfrac{4}{s}}{1 + \dfrac{3}{s}}$

$\mathcal{L} \left\{ f(t) \right\} = \ln \dfrac{1 - 0}{1 + 0} - \ln \dfrac{\dfrac{s - 4}{s}}{\dfrac{s + 3}{s}}$

$\mathcal{L} \left\{ f(t) \right\} = \ln 1 - \ln \dfrac{s - 4}{s + 3}$

$\mathcal{L} \left\{ f(t) \right\} = 0 - [ \, \ln (s - 4) - \ln (s + 3) \, ]$

$\mathcal{L} \left\{ f(t) \right\} = \ln (s + 3) - \ln (s - 4)$

$\mathcal{L} \left\{ f(t) \right\} = \ln \left( \dfrac{s + 3}{s - 4} \right)$           answer