Chapter 10 - The Laplace Transform and Some Elementary Applications - 10.1 Definition of the Laplace Transform - Problems - Page 675: 6

$\dfrac{s}{(s^2-b^2) }$

The Laplace Transform can be written as: $L[F(t)]=\int_{0}^{\infty} e^{-st} f(t) dt $ We are given that $f(t)=\cos (h bt)$ Now, $L[F(t)]=\int_{0}^{\infty} e^{-st} f(t) dt \\= \int_{0}^{\infty} e^{-st} [\cos (h bt)] \ dt \\=\int_{0}^{\infty} \dfrac{e^{-(s-b)t} \ dt}{2}+\int_{0}^{\infty} \dfrac{e^{-(s+b)t} \ dt}{2} \\=\dfrac{1}{2(s-b) }-\dfrac{1}{2(s+b)} \\=\dfrac{s}{(s^2-b^2) }$