Asymptotic expansion of $J(t) = \int^{\infty}_{0}{\exp(-t(x + 4/(x+1)))}\, dx$ I want to derive an asymptotic expansion for the following Bessel function. I think I need to rewrite it in another form, from which I can integrate it by parts. I am interested in obtaining the expansion up to second order as $t$ approaches infinity.
$$J(t) = \int^{\infty}_{0}{\exp\left(-t\left(x + \dfrac{4}{x+1} \right) \right)}\, dx$$

Please note that the leading exponential terms of your integral $J(t)$ can be expressed in terms of standard "modified Bessel functions" whose asymptotic expansion is known.
Let $z = \frac{x+1}{2} = e^\theta$, we have
$$\begin{align}
J(t) 
& = e^t \int_0^\infty e^{-2t (\frac{x+1}{2} + \frac{2}{x+1})} dx
= 2 e^t \int_{\frac12}^\infty e^{-2t (z + z^{-1})} dz\\
& = 2 e^t \left[ \left( \int_0^\infty - \int_0^{\frac12} \right) e^{-2t (z + z^{-1})} dz \right]\\
& = 2 e^t \left[ \int_0^\infty e^{-2t (z + z^{-1})} dz - O\left( \int_0^{\frac12}e^{-2t(\frac12 + 2)} dz\right)\right]\\
& = 2e^t\int_{-\infty}^\infty e^{-4t\cosh\theta} e^\theta d\theta - O( e^{-4t} )\\
& = 4e^t\int_0^\infty e^{-4t\cosh\theta} \cosh\theta d\theta - O(e^{-4t})\\
& = 4e^t K_1(4t) - O(e^{-4t})
\end{align}
$$
where $K_\alpha(z)$ is the modified Bessel function
 of the $2^{nd}$ kind. Using following asymptotic expansion of $K_\alpha(z)$:
$$K_\alpha(z) \sim \sqrt{\frac{\pi}{2z}} e^{-z} \left( 1 + \sum_{m=1}^{\infty} \frac{\prod_{k=1}^m (4\alpha^2 - (2k-1)^2)}{m!(8z)^m}\right)$$
We get 
$$\begin{align}
J(t) \sim & \sqrt{\frac{2\pi}{t}} e^{-3t} \left(1 + \sum_{m=1}^{\infty} \frac{\prod_{k=1}^m (4 - (2k-1)^2)}{m!(8z)^m}\right) - O(e^{-4t})\\
= & \sqrt{\frac{2\pi}{t}} e^{-3t} \left( 1 + 
\frac{3}{32t}-\frac{15}{2048 t^2}+\frac{105}{65536 t^3}-\frac{4725}{8388608 t^4} 
+ \cdots \right) - O(e^{-4t})
\end{align}$$
Update
There are several ways to compute the asymptotic expansion of $J(t)$. 
Other answers has shown how to cast it to an integral of the form 
$$J(t) = \int e^{-t\varphi(y)} dy$$
and then expand $\varphi(y)$ to get desired result. One can also rewrite $J(t)$ 
to the form:
$$J(t) = \int_0^\infty e^{-ty}\psi(y) dy$$
and uses Watson's Lemma
to obtain the asymptotic expansion. 
Let I use $K_1(z)$ as an example. 
In the integral representation of $K_1(z)$, let $1 + \eta = \cosh\theta$, we have
$$\begin{align}
K_1(z) & = \int_0^\infty e^{-z\cosh\theta} \cosh\theta d\theta\\
& = \int_0^\infty e^{-z(1+y)} \frac{1+y}{\sqrt{(y+1)^2-1}} dy\\
& = \frac{e^{-z}}{\sqrt{2}} \int_0^\infty e^{-zy}\frac{1+y}{\sqrt{y(1 + \frac{y}{2}})} dy\\
& \stackrel{formal}{=} \frac{e^{-z}}{\sqrt{2}} \int_0^\infty e^{-zy}\frac{1+y}{\sqrt{y}}
\left[ 1 + \sum_{m=1}^\infty \frac{(-1)^m}{m!} \left(\frac12 \right)_m \left(\frac{y}{2} \right)^m \right] dy\\
& = \frac{e^{-z}}{\sqrt{2}} \int_0^\infty \frac{e^{-zy}}{\sqrt{y}}
\left[ 1 + \sum_{m=1}^\infty \frac{(-1)^m}{2^m m!} 
\left(\left(\frac12\right)_m - 2m\left(\frac12\right)_{m-1}\right)
y^m \right] dy\\
& = \frac{e^{-z}}{\sqrt{2}} \int_0^\infty \frac{e^{-zy}}{\sqrt{y}}
\left[ 1 + \sum_{m=1}^\infty \frac{(-1)^{m-1}}{2^m m!} 
\left(\left(\frac12\right)_{m-1} (\frac12+m) \right)
y^m \right] dy\\
\end{align}$$
where $(\alpha)_m = \alpha(\alpha+1)\cdots(\alpha+m-1)$ is the rising 
Pochhammer symbol.
In above transformations of the integral, the steps after the $\stackrel{formal}{=}$ 
are formal manipulations of equations. This is because the power series in
the integrand has a finite radius of convergence. However, this doesn't stop
the last expression we arrived from being useful. Watson's Lemma tell us if
the factor inside the $\left[ \cdots \right]$ doesn't grow too fast,  then we
can formally integrate the expression term by term and read off the asymptotic expansion of the integral directly. In this case, we have:
$$\begin{align}
K_1(z) 
& \sim 
\frac{e^{-z}}{\sqrt{2}} \frac{\Gamma(\frac12)}{\sqrt{z}}
\left[ 1 + \sum_{m=1}^\infty \frac{(-1)^{m-1}}{2^m m!} 
\left(\left(\frac12\right)_{m-1} (\frac12+m) \frac{\Gamma(\frac12+m)}{\Gamma(\frac12)} \right)
\frac{1}{z^m} \right]\\
& = \sqrt{\frac{\pi}{2z}} e^{-z}\left[
1 + \sum_{m=1}^\infty \frac{(-1)^{m-1}}{2^m m!}
\left(\frac12\right)_{m-1} \left(\frac12\right)_{m+1} \frac{1}{z^m}
\right]\\
& = \sqrt{\frac{\pi}{2z}} e^{-z}\left[
1 + \sum_{m=1}^\infty \frac{\prod_{k=1}^m (4 - (2k-1)^2)}{8^m m!} 
\frac{1}{z^m}
\right]
\end{align}
$$
If one really want, one can compute the asymptotic expansion of the $O(e^{-4t})$ term in $J(t)$ using same method.
