$f(t) = \dfrac{\cos 4t - \cos 5t}{t}$
$\mathcal{L} \left\{ f(t) \right\} = \mathcal{L} \left[ \dfrac{\cos 4t - \cos 5t}{t} \right]$
$\mathcal{L} \left\{ f(t) \right\} = \mathcal{L} \left[ \dfrac{\cos 4t}{t} \right] - \mathcal{L} \left[ \dfrac{\cos 5t}{t} \right]$
Since
$\mathcal{L} (\cos bt) = \dfrac{s}{s^2 + b^2}$
Then,
$\displaystyle \mathcal{L} \left\{ f(t) \right\} = \int_s^\infty \dfrac{u \, du}{u^2 + 4^2} - \int_s^\infty \dfrac{u \, du}{u^2 + 5^2}$
$\displaystyle \mathcal{L} \left\{ f(t) \right\} = \int_s^\infty \left( \dfrac{u}{u^2 + 16} - \dfrac{u}{u^2 + 25} \right) \, du$
$\displaystyle \mathcal{L} \left\{ f(t) \right\} = \dfrac{1}{2}\int_s^\infty \left( \dfrac{2u}{u^2 + 16} - \dfrac{2u}{u^2 + 25} \right) \, du$
$\mathcal{L} \left\{ f(t) \right\} = \frac{1}{2} \Big[ \ln (u^2 + 16) - \ln (u^2 + 25) \Big]_s^\infty$
$\mathcal{L} \left\{ f(t) \right\} = \frac{1}{2} \left[ \ln \dfrac{u^2 + 16}{u^2 + 25} \right]_s^\infty$
$\mathcal{L} \left\{ f(t) \right\} = \frac{1}{2} \left[ \ln \dfrac{\dfrac{u^2}{u^2} + \dfrac{16}{u^2}}{\dfrac{u^2}{u^2} + \dfrac{25}{u^2}} \right]_s^\infty$
$\mathcal{L} \left\{ f(t) \right\} = \frac{1}{2} \left[ \ln \dfrac{1 + \dfrac{16}{u^2}}{1 + \dfrac{25}{u^2}} \right]_s^\infty$
$\mathcal{L} \left\{ f(t) \right\} = \frac{1}{2} \left[ \ln \dfrac{1 + \dfrac{16}{\infty^2}}{1 + \dfrac{25}{\infty^2}} - \ln \dfrac{1 + \dfrac{16}{s^2}}{1 + \dfrac{25}{s^2}} \right]$
$\mathcal{L} \left\{ f(t) \right\} = \frac{1}{2} \left[ \ln \dfrac{1 + 0}{1 + 0} - \ln \dfrac{\dfrac{s^2 + 16}{s^2}}{\dfrac{s^2 + 25}{s^2}} \right]$
$\mathcal{L} \left\{ f(t) \right\} = \frac{1}{2} \left[ \ln 1 - \ln \dfrac{s^2 + 16}{s^2 + 25} \right]$
$\mathcal{L} \left\{ f(t) \right\} = \frac{1}{2} \Big\{ 0 - [ \, \ln (s^2 + 16) - \ln (s^2 + 25) \, ] \Big\}$
$\mathcal{L} \left\{ f(t) \right\} = \frac{1}{2} \Big\{ \ln (s^2 + 25) - \ln (s^2 + 16) \Big\}$
$\mathcal{L} \left\{ f(t) \right\} = \frac{1}{2} \ln \dfrac{s^2 + 25}{s^2 + 16}$ answer
