Answer
See below
Work Step by Step
Since the given function is periodic on $[0,∞)$ with period $T=1$, we have
$L(f)=\frac{1}{1-e^{-s}}\int ^1_0e^{-st}f(t)dt=\frac{1}{1-e^{-s}}\int ^1_0 t e^{-st} dt$
Using the standard integral
$\int e^{at}\sin bt dt=\frac{1}{a^2+b^2}e^{at}(a\sin bt-b\cos bt)+c$
it follows that
$L(f)=\frac{1}{1-e^{-s}}\int ^1_0te^{-st} dt\\
=\frac{1}{1-e^{-2s}}[\frac{-e^{-s}}{s}+\frac{1-e^{-s}}{s^2}]\\
=\frac{1}{1-e^{-2s}}[\frac{(1-e^{-s})(1+s)}{s^2}]$
