How to determine the leading order asymptotic behaviour of this integral? I tried to put it in the form f(t) exp(tx) to apply watson's lemma but I couldn't. Is there a trick?
$$\int_{0}^{1} e^{t}\left ( \frac{t}{1+t^{2}} \right )^{x} dt$$  as $ x \rightarrow {\infty} $
 A: $$I_x=\int_{0}^{1} e^{t}\left ( \frac{t}{1+t^{2}} \right )^{x} dt$$
$$\frac{t}{1+t^{2}}=\frac 1 2 \left(\frac 1 {t+i}+\frac 1 {t-i} \right)$$
$$I_x=\frac 1 {2^x}\int_{0}^{1} e^{t}\left(\frac 1 {t+i}+\frac 1 {t-i} \right)^x \,dt$$ So, for an integer value of $x$, you face linear combinations of exponential integral functions of conjugate complex arguments.
For example
$$I_3= \frac {1-e}4+\frac {A_3-B_3}{16}$$
$$A_3=(-1+3 i)\, e^{+i}\, \big[\text{Ei}(-i)-\text{Ei}(1-i)\big]$$
$$B_3=(+1+3 i) \,e^{-i} \, \big[\text{Ei}(+i)-\text{Ei}(1+i)\big]$$ Working the $I_x$, you could easily find interesting patterns.
As a result, for large values of $x$
$$I_x \sim \frac K{2^x}\qquad \text{with} \qquad K\sim \frac 1 3\color{red}{\text{    Wrong}}$$
Computing exactly for $x=100$, the value is $2.32\times 10^{-21}$ while the approximation gives $2.63\times 10^{-21}$.
Edit
After @Maxim's comments and @mathstackuser12's edit, I checked again my calculations and found some severe comptational errors.
I just performed a quick and dirty nonlinear regression for the model
$$I_x= a\, x^b\, 2^{-x}$$ for the range $5 \leq x \leq 1000$ (step=$5$) and obtained with $R^2=0.998378$
$$\begin{array}{clclclclc}
 \text{} & \text{Estimate} & \text{Standard Error} & \text{Confidence
   Interval} \\
 a & +1.88546 & 0.01574 & \{+1.85441,+1.91650\} \\
 b & -0.41209 & 0.00175 & \{-0.41555,-0.40864\} \\
\end{array}$$
A: Let
$$S\left( x \right)=\int\limits_{0}^{1}{{{e}^{t}}{{e}^{x\log \left( \frac{t}{1+{{t}^{2}}} \right)}}dt}=\int\limits_{0}^{1}{{{e}^{t}}{{e}^{xh\left( t \right)}}dt}$$
Note that
$$h'\left( {{t}_{0}} \right)=0\Rightarrow t=1$$
And so the end point is a stationary point of the argument against x, therefore consider
$$S\left( x \right)=\int\limits_{0}^{1}{{{e}^{t}}{{e}^{x\left( h\left( 1 \right)+\frac{1}{2}h''\left( 1 \right){{\left( t-1 \right)}^{2}}+... \right)}}dt}$$
Taking an approximation about the maximum and keeping only the first term of ${e}^{t}$ and the quadratic term in $h\left(t\right)$
$$S\left( x \right)\simeq {{e}^{xh\left( 1 \right)+1}}\int\limits_{1-\delta }^{1}{{{e}^{\tfrac{1}{2}xh''\left( 1 \right){{\left( t-1 \right)}^{2}}}}dt}\simeq {{e}^{xh\left( 1 \right)+1}}\int\limits_{-\delta }^{0}{{{e}^{\tfrac{1}{2}xh''\left( 1 \right){{u}^{2}}}}du}\simeq {{e}^{-x\log \left( 2 \right)+1}}\int\limits_{-\infty }^{0}{{{e}^{-\tfrac{1}{2}x{{u}^{2}}}}du}$$
Completing the integral we have (effectively by Watson’s lemma)
$$S\left( x \right)\sim \sqrt{\frac{\pi }{2x}}{{e}^{-x\log \left( 2 \right)+1}}=\sqrt{\frac{\pi }{2x}}\frac{e}{{{2}^{x}}}$$
