The basic idea is that the maximum contribution of the integral comes from a neighborhood of $t=0$, and near there we have $t^2+2t^4 \approx t^2$. This problem is particularly nice because we can do everything explicitly. I'll do the calculation in two steps (two changes of variables) to illustrate what's going on.
Start with the change of variables $t^2+2t^4 = s^2$, where $s \geq 0$. This gives
$$
t=\frac{1}{2}\sqrt{-1+\sqrt{1+8s^2}},
$$
so that the integral becomes
$$
\int_0^{\infty} t^{3/4}e^{-x(t^2+2t^4)}\,dt = \int_0^\infty s^{3/4} f(s) e^{-xs^2}\,ds,
$$
where
$$
f(s) = \frac{2^{1/4}s^{1/4}}{\sqrt{1+8s^2}\left(-1+\sqrt{1+8s^2}\right)^{1/8}} = 1 - \frac{15}{4}s^2 + \frac{713}{32}s^4 + \cdots.
$$
Now we can make the second change of variables $s^2 = r$ to put the integral into a form where we can directly apply Watson's lemma. Indeed, this gives
$$
\int_0^\infty s^{3/4} f(s) e^{-xs^2}\,ds = \int_0^\infty r^{-1/8} g(r) e^{-xr}\,dr,
$$
where
$$
g(r) = \frac{1}{2}f\left(\sqrt{r}\right) = \frac{1}{2} - \frac{15}{8}r + \frac{713}{64}r^2 + \cdots.
$$
Finally
$$
\begin{align*}
\int_0^\infty r^{-1/8} g(r) e^{-xr}\,dr &\approx \sum_{n=0}^{\infty} \frac{g^{(n)}(0) \Gamma(n+7/8)}{n! x^{n+7/8}} \\
&= \frac{1}{2}\Gamma\left(\frac{7}{8}\right) x^{-7/8} - \frac{15}{8}\Gamma\left(\frac{15}{8}\right)x^{-15/8} + O\left(x^{-23/8}\right)
\end{align*}
$$
as $x \to \infty$, by Watson's lemma.
Best Answer
If $t\ge 1$, the integral diverges. If $0<t<1$, we can write \begin{align*} \int_0^{ + \infty } {t^{u + \frac{1}{u}} u^k {\rm d}u} & = \int_0^{ + \infty } {{\rm e}^{\left( {u + \frac{1}{u}} \right)\log t} u^k {\rm d}u} \\ & = \left( {\frac{{2\left| {\log t} \right|}}{2}} \right)^{ - (k + 1)} \int_0^{ + \infty } {{\rm e}^{ - s - \frac{{(2\left| {\log t} \right|)^2 }}{{4s}}} \frac{{{\rm d}s}}{{s^{ - (k + 1) + 1} }}} \\ & = 2K_{ - (k + 1)} (2\left| {\log t} \right|) = 2K_{k + 1} (2\left| {\log t} \right|), \end{align*} where $K_\nu$ is the modified Bessel function of the second kind (cf. $(10.32.10)$). Then, by $\S10.40(\text{iii})$ in the DLMF, $$ 2K_{k + 1} (2\left| {\log t} \right|) \le \sqrt {\frac{\pi }{{\left| {\log t} \right|}}} t^2 \left( {1 + \frac{{4(k + 1)^2 - 1}}{{8\left| {\log t} \right|}}\exp \left( {\frac{{4(k + 1)^2 - 1}}{{8\left| {\log t} \right|}}} \right)} \right) $$ for any $0<t<1$, and this is asymptotically sharp as $t\to 0^+$.