I got a result that doesn't seem correct, so I'm hoping someone can tell me if I went wrong somewhere (probably with the integration by parts or in the second to last line).
$$ L(x(t)^2x'(t))=\int_{0}^{\infty} x(t)^2x'(t)e^{-st}dt $$
So I use integration by parts with
$$ u=x(t)^2 $$ $$ dv=x'(t)e^{-st}dt $$
Now my guess is this is where I went wrong and somehow misused the identity for the transform of the derivative of the function:
$$ du=2x(t)x'(t)dt $$ $$ v=sX(s)-x(0) $$
Then
$$ L(x(t)^2x'(t))=[sX(s)-x(0)][x(t)^2|_{0}^{\infty}-\int 2x(t)x'(t)dt] $$ $$ =[sX(s)-x(0)][x(t)^2|_{0}^{\infty}-\int 2x(t)dx(t)] $$ $$ =[sX(s)-x(0)][x(t)^2|_{0}^{\infty}-x(t)^2|_{0}^{\infty}] $$ $$ =0 $$
If this result is indeed correct then that's a pretty shocking identity, otherwise hopefully someone can point out where I went wrong.
Thanks.
You integrate $dv$ the wrong way. Choose $u'=x'(t) x(t)^2$ and $v=e^{-st}$. Note that $u=\frac{x(t)^3}{3}$.