Chapter 10 - The Laplace Transform and Some Elementary Applications - 10.7 The Second Shifting Theorem - Problems - Page 704: 1

$\dfrac{e^{-2s}-e^{-3s}}{s}$

$L (u_a(t)) =\int_0^{\infty} e^{-st} u_a(t) dt \\=\int_0^{\infty} e^{-st} dt\\=[\dfrac{e^{-st}}{-s}]_0^{\infty}$ Now, $L( u_2(t)-u_3(t)]=\dfrac{e^{-2s}}{s}-\dfrac{e^{-3s}}{s}\\=\dfrac{e^{-2s}-e^{-3s}}{s}$