Question: Verify:

$$y\left(\frac{1}{2}\right) = e^\frac{1}{2} \displaystyle \int\_\frac{1}{2}^{\infty} \frac{dt}{te^t} ≈ 0.9232.$$

I've written the integral as: 

$$\lim\_{x \to \infty} \displaystyle\int\_\frac{1}{2}^x \frac{dt}{te^t}$$

I presume I should use substitution to continue but not sure why or how.

We have gone over Trapezoidal and Simpson's Rule for numerical integration so I presume that's what I should use to solve it

Answer: 
Notice that

\begin{align\*}
\int\_{12}^{\infty} \frac{1}{te^t} \, dt &\le \sum\_{n = 12}^{\infty} \frac{1}{ne^n} \\
&\le \frac 1 {12e^{12}} \sum\_{n = 0}^{\infty} \frac 1 {e^n} \\
&< \frac{1}{e^{12}} \\
&< 0.000006
\end{align\*}

is already small enough that it won't affect your computation's first four decimal places. Now just compute $\int\_{1/2}^{12}$ using your favorite numerical method. (And by the way, this was a very crude method of estimate, so $12$ is a very loose upper bound.)
We have $$\mathfrak{I}=e^{1/2}\int\_{1/2}^{+\infty}\frac{dx}{x e^x}\,dx = \int\_{0}^{+\infty}\frac{dt}{\left(t+\tfrac{1}{2}\right)e^t}=\int\_{0}^{1}\frac{du}{\tfrac{1}{2}-\log u}=\int\_{0}^{1}\frac{v}{\tfrac{1}{4}-\log v}\,dv$$
where the last form is well-suited for standard techniques: the integration range is a bounded interval and the integrand function has bounded derivatives over the closed interval. For instance, through the trapezoidal rule:
$$ \mathfrak{I} \approx \frac{1}{100}\left[2+\sum\_{k=1}^{99}\frac{\tfrac{k}{100}}{\tfrac{1}{4}-\log\tfrac{k}{100}}\right] \approx 0.923. $$