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.
This is solvable by Laplace's method or the saddle point approximation.
If $t$ becomes large only those parts of the integrand where $f(x) = x + \frac{4}{x+1}$ is small will have a relevant contribution.
So one searches the minima of $f(x)$. In this case there is only one: $x_0 = 1$. One now approximates $f(x)$ by its Taylor expansion up to second order around $x_0$. The first order term will vanish since this is a minimum and the second order term will be positive, otherwise $x_0$ would not be a minimum but a maximum. In this case we arrive at:
$$f(x) = 3 + \frac{1}{2}(x-1)^2 + \mathcal{O}(x^3)$$
and with that, ignoring the higher order terms:
$$ J(t) = \int\limits_{0}^\infty dx\; e^{-3t - \frac{t}{2}(x-1)^2} = e^{-3t}\int\limits_0^\infty dx\; e^{-\frac{1}{2}(x-1)^2} = \sqrt{\frac{2\pi}{t}} e^{-3t} $$
If you want to refine the solution, you can consider the remainder in the Taylor expansion as a perturbation. Let $g(x) = f(x) - 3 - \frac{1}{2}(x-1)^2$ be the error we made previously. Then you can approximate $e^{-tg(x)}$ around $x_0$ as $1-tg(x)$, because for large $t$ its contribution will vanish (there is no zeros order term in $g(x)$).
So you have: $$ J(t) = e^{-3t}\int\limits_0^\infty dx\; e^{-\frac{t}{2}(x-1)^2}\left(1-\frac{t}{6}g'''(1)(x-1)^3 - \frac{t}{24}g''''(1)(x-1)^4 + \ldots\right) $$
The integrals for every summand can be solved. However they are not that nice. You may notice that all terms with $x$ to an odd power will vanish because a term exponential in $t$ will remain. All even powers will reproduce $\sqrt{\frac{2\pi}{t^k}}$ with a constant factor for some $k$, resulting in the expression given by Raymond Manzoni.
First make the (cosmetic) change of variables $x = y+1$, so that
$$ \begin{align} J(t) &= \int^{\infty}_{0} \exp\left[-t\left(x + \dfrac{4}{x+1} \right) \right]\, dx \\ &= \int_{-1}^{\infty} \exp\left[-t\left(y + 1 + \dfrac{4}{y+2} \right) \right]\, dy \\ &= \int_{-1}^{\infty} \exp\Bigl[-t f(y) \Bigr]\,dy. \end{align} $$
Now $f(y)$ has a minimum at $y=0$, and near there we have
$$ f(y) = 3 + \frac{y^2}{2} - \frac{y^3}{4} + \frac{y^4}{8} + \cdots. $$
We would like to introduce a new variable by
$$ 3 + \frac{z^2}{2} = 3 + \frac{y^2}{2} - \frac{y^3}{4} + \frac{y^4}{8} + \cdots $$
or, taking the principal branch of the square root,
$$ z = y \sqrt{1 - \frac{y}{2} + \frac{y^2}{4} + \cdots}, $$
which of course would only hold in a small neighborhood of the origin. In general we can solve for $y$ in this equation using series reversion (the Lagrange formula, say), but in this case we can write down the answer explicitly as
$$ \begin{align} y &= \frac{1}{4} \left(z^2 + z \sqrt{16+z^2}\right) \\ &= z + \frac{z^2}{4} + \frac{z^3}{32} - \frac{z^5}{2048} + O\left(z^7\right). \end{align} $$
Thus we have
$$ dy = \left[1+\frac{z}{2}+\frac{3 z^2}{32}-\frac{5 z^4}{2048} + O\left(z^6\right)\right]dz. $$
The integrals over the tails of the intervals are exponentially small as $t \to \infty$, and by following the usual steps in the Laplace method we arrive at the expression
$$ J(t) = \int_{-\infty}^{\infty} \exp\left[-t \left(3 + \frac{z^2}{2}\right)\right] \left(1+\frac{z}{2}+\frac{3 z^2}{32}-\frac{5 z^4}{2048}\right)\,dz + O\left(\int_{-\infty}^{\infty} \exp\left[-t \left(3 + \frac{z^2}{2}\right)\right] z^6\,dz\right). $$
We can integrate term-by-term and conclude that
$$ J(t) = e^{-3t} \sqrt{\frac{2\pi}{t}} \left[1 + \frac{3}{32} t^{-1} - \frac{15}{2048} t^{-2} + O\left(t^{-3}\right)\right] $$
as $t \to \infty$.
