= Solution
Put
$$
\phi(t)=t^3-2t^2+t=t(t-1)^2.
$$
On $[0,2]$, its global maximum is the endpoint $t=2$, where
$$
\phi(2)=2,
\qquad
\phi'(2)=5,
$$
and the amplitude $\log t$ has value $\log2$. The endpoint form of <Laplace's method> therefore gives
$$
\boxed{
I(x)\sim\frac{\log2}{5x}e^{2x}
}.
$$
Indeed, with $u=2-t$, Taylor expansion gives $\phi(2-u)=2-5u+O(u^2)$ and $\log(2-u)=\log2+O(u)$; the scale contributing to the integral is $u=O(x^{-1})$. Outside any fixed neighbourhood of $2$, $\phi$ is bounded above by $2-\delta$, so that part is exponentially smaller. This also controls the integrable logarithmic singularity at zero.
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