Compute the Laplace transform of f(t)={(8 ,if 0 <= x <14),(9t,if 14 <= x <infty):}

Step 1
Consider the given information.
$f(t)=\{\begin{array}{ll}8& \text{if }0\le x<14\\ 9t& \text{if }14\le x<\mathrm{\infty }\end{array}$
The Laplace transform is defined as,
${\displaystyle F\left(s\right)={\int }_0^{\mathrm{\infty }}{e}^{-st}f\left(t\right)dt}$
Put the value in the integral.
$L\left\{f\right\}(s)={\int }_0^{14}{e}^{-st}8dt+{\int }_{14}^{\mathrm{\infty }}{e}^{-st}(9t)dt$
${\displaystyle =8{\int }_0^{14}{e}^{-st}dt+9{\int }_{14}^{\mathrm{\infty }}t{e}^{-st}dt}$
Step 2
Consider the values for integration as,
${\displaystyle u=t,dv=e^{-st}}$
${\displaystyle du=1,v=\frac{e^{-st}}{-s}}$
Put the values in the function.
$L\left\{f\right\}(s)=8{\int }_0^{14}{e}^{-st}+9{\int }_{14}^{\mathrm{\infty }}t{e}^{-st}$
${\displaystyle =8{\left[\frac{e^{-st}}{-s}\right]}_0^{14}+9{\left[\frac{t{e}^{-st}}{-s}\right]}_{14}^{\mathrm{\infty }}+\frac{1}{s}{\int }_{14}^{\mathrm{\infty }}{e}^{-st}dt}$
${\displaystyle =-8\frac{(e^{-14s}-1)}{s}-9{\left[\frac{t{e}^{-st}}{s}\right]}_{14}^{\mathrm{\infty }}-9\frac{1}{s}{\left[\frac{e^{-st}}{s}\right]}_{14}^{\mathrm{\infty }}}$
${\displaystyle =-8\frac{(e^{-14s}-1)}{s}+\frac{(126s+9)e^{-14s}}{s^2}}$
${\displaystyle =\frac{118e^{-14s}s+8s+9e^{-14s}}{s^2}}$