Find the laplace transform of the following: MULTIPLICATION BY POWER OF t g(t)=(t2−3t+2)sin⁡(3t)

Step 1
To find Laplace transform of $g(t)=(t^2-3t+2)\sin (3t)$, we simplify the expression and we get
$g(t)=t^2\sin (3t)-3t\sin (3t)+2\sin (3t)$
Using the following results:
$1.L(f(t))=F(s)$
$2.L(t^{n}f(t))=(-1{)}^n\frac{d^{n}F(s)}{d{s}^n}$
$3.L(\sin (at))=\frac{a}{s^2+a^2}$
Step 2
Using the above results and finding the required transforms,
$L(\sin (3t))=\frac{3}{s^2+3^2}=\frac{3}{s^2+9}$
$L(t^2\sin (3t))=(-1)\frac{2}{d}$
Now,
$L(-3t\sin (3t))=-3(-1)\frac{d}{ds}(L(\sin (3t)))$
$=3\frac{d}{ds}\left(\frac{3}{s^2+9}\right)$
$=3\left(\frac{-3\cdot 2s}{(s^2+9{)}^2}\right)$
$=\frac{-18s}{(s^2+9{)}^2}$
Now,
$L(2\sin (3t))=2\frac{3}{(s^2+9)}$
$=\frac{6}{(s^2+9)}$
Step 3
Therefore,
$L(g(t))=L(t^2\sin (3t)-3t\sin (3t)+2\sin (3t))$
$=L[t^2\sin (3t)]+L[-3t\sin (3t)]+L[2\sin (3t)]$
Substituting the values of Laplace transforms from above step, we get
$G(s)=\frac{6[3s^2-9]}{(s^2+9{)}^3}-\frac{18s}{(s^2+9{)}^2}+\frac{6}{(s^2+9)}$