$\displaystyle L\left[\cos{ax}\right]=\frac{s}{s^2+a^2} \\[2ex] \begin{align*} \displaystyle L\left[\cos{6t\ }-\cos{4t}\right]&=\ \frac{s}{s^2+6^2}-\frac{s}{s^2+4^2}\\[2ex]&=\ \frac{s}{s^2+36}-\frac{s}{s^2+16} \\[2ex] \end{align*}$

$ \displaystyle Now,\ L\left[\frac{f\left(x\right)}{x}\right]=\int_s^{\infty{}}F(s)ds \\[2ex] \displaystyle \therefore{},\ L\left[\frac{\cos{6t}-\cos{4t}}{t}\right]\\[2ex] \displaystyle=\ \int_s^{\infty{}}\left(\frac{s}{s^2+36}-\frac{s}{s^2+16}\right)ds\\[3ex] \displaystyle =\frac{1}{2}\int_s^{\infty{}}\left(\frac{2s}{s^2+36}-\frac{2s}{s^2+16}\right)ds \\[2ex] \displaystyle =\frac{1}{2}{\left[\ln{\left(s^2+36\right)}-\ln{\left(s^2+16\right)}\right]}_s^{\infty{}} \\[2ex] \displaystyle =\frac{1}{2}{\left[\ln{\frac{s^2+36}{s^2+16}}\right]}_s^{\infty{}} \\[2ex] \displaystyle =\frac{1}{2}{\left[\ln{\frac{1+\frac{36}{s^2}}{1+\frac{16}{s^2}}}\ \right]}_s^{\infty{}}\\[2ex] \displaystyle =\frac{1}{2}\left[\ln{\frac{1+0}{1+0}}-\ln{\frac{s^2+36}{s^2+16}}\right]\\[2ex] \displaystyle =\frac{1}{2}\left[0-\ln{\frac{s^2+36}{s^2+16}}\right] \\[2ex] \displaystyle =\frac{-1}{2} \ln{\frac{s^2+36}{s^2+16}} \\[2ex] $

$From \ the \ definition \ of \ Laplace \ Transform,\\[2ex] \displaystyle L\left[f\left(t\right)\right]=\int_0^{\infty{}}f\left(t\right)e^{-st}dt \\[2ex] \displaystyle L\left[\frac{\cos{6t}-\cos{4t}}{t}\right]=\int_0^{\infty{}}\frac{\cos{6t\ }-cos\ 4t}{t}e^{-st}dt\ \\[3ex] \displaystyle \frac{-1}{2}\ln{\frac{s^2+36}{s^2+16}}=\int_0^{\infty{}}\frac{\cos{6t\ }-\cos\ 4t}{t}e^{-st}dt \\[2ex] $

$\displaystyle Plug\ in\ s=0, \\[2ex] \displaystyle \int_0^{\infty{}}\frac{\cos{6t\ }-\cos\ 4t}{t}e^{-0t}dt=\frac{-1}{2}\ln{\frac{0+36}{0+16}}\ \ \\[4ex] $

$\displaystyle \int_0^{\infty{}}\frac{\cos{6t\ }-\cos 4t}{t}dt=\ -\frac{1}{2}\ln{{\left(\frac{6}{4}\right)}^2\\[2ex]=\ }-\frac{2}{2}\ln{\frac{3}{2}}\\[2ex] =\ln{\dfrac{2}{3}} \\[2ex] $