$L[\cos at] = \dfrac{s}{s^2+a^2} \ L[\cos 3t -\cos 2t] = \dfrac{s}{s^2+ 3^2} - \dfrac{s}{s^2+2^2 } $ $\ Now, \: L[\dfrac{f(t)}{t}] = \int \limits_s^\infty F(s)ds \ \therefore, L[\dfrac{\cos 3t -\cos 2t}{t}]\ = \int \limits_s^\infty [\dfrac{s}{s^2+ 3^2} - \dfrac{s}{s^2+2^2 }] ds $ $\ =\dfrac{1}{2} \int \limits_s^\infty [\dfrac{2s}{s^2+ 9}- \dfrac{2s}{s^2+ 4}]ds \ =\dfrac{1}{2} [\ln(s^2+9)-\ln(s^2+4)]_s^\infty $ $ \ =\dfrac{1}{2} [\ln \dfrac{s^2+9}{s^2+4}]_s^\infty \= \dfrac{1}{2} [\ln \dfrac{1+\dfrac{9}{s^2}}{1+\dfrac{4}{s^2}}]_s^\infty $ $\ =\dfrac{1}{2} [\ln\dfrac{1+0}{1+0} -\ln \dfrac{s^2+9}{s^2+4}] \ = -\dfrac{1}{2} [\ln\dfrac{s^2+9}{s^2+4}] ...(A)$ $\text{From the definition of Laplace transform,} \ \int \limits_0^\infty f(t)e^{-st}dt = L[f(t)]$ $ \ \int \limits_0^\infty e^{-st}(\dfrac{\cos 3t-\cos 2t}{t})dt = L[\dfrac{\cos 3t-\cos 2t}{t}] \ \int \limits_0^\infty e^{-st}(\dfrac{\cos 3t-\cos 2t}{t})dt = -\dfrac{1}{2}[\ln \dfrac{s^2+9}{s^2+4}] ... from (A) $ $\ \text{Plug in s=1} \ \int \limits_0^\infty e^{-t}(\dfrac{\cos 3t-\cos 2t}{t})dt = -\dfrac{1}{2}[\ln \dfrac{1^2+9}{1^2+4}]= -\dfrac{1}{2}[\ln \dfrac{10}{5}] \ \bbox[3pt, yellow]{\therefore, \int \limits_0^\infty e^{-t}(\dfrac{\cos 3t-\cos 2t}{t})dt =-\dfrac{1}{2}\ln2}$