Calculate the Laplace transform L\{f\} of the function f(t)=e^{4t+2}-8t^6+8 \sin (2t)+9 using the basic formulas

Step 1
${\displaystyle f\left(t\right)=e^{4t+2}-8t^6+8\sin \left(2t\right)+9}$
${\displaystyle F\left(t\right)=e^2{e}^{4t}-8t^6+8\sin \left(2t\right)+9}$
$L\{f(t)\}=L\{e^2{e}^{4t}-8t^6+8\sin (2t)+9\}$
Apply linearity
$L\{f(t)\}=e^{2}L\left\{e^{4t}\right\}-8L\left\{t^6\right\}-8L\{\sin (2t)\}+9L\left\{1\right\}$
using the basic formulas
$L\left\{e^{at}\right\}=\frac{1}{s-a},L\left\{1\right\}=\frac{1}{s}$
$L\left\{t^n\right\}=\frac{n!}{s^{n+1}}$
and $L\{\sin (at)\}=\frac{a}{s^2+a^2}$
So, $L\{f(t)\}=e^2\frac{1}{s-4}-8\frac{6!}{s^7}-8\frac{2}{s^2+2^2}+9\frac{1}{s}$
$L\left\{f\right\}=\frac{e^2}{s-4}-\frac{5760}{s^7}-\frac{16}{s^2+2^2}+\frac{9}{s}$