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.)
Since the boundary does not enclose zero, change the integral variable $x=\frac{1}{t}$. Then the better method will be Gaussian quadratures to solve the integral numerically.

Assume $x=1/t$ then the integral becomes

$$\int\_0^2\dfrac{e^{-1/x}}{x}dx$$

then standardize it by defining $x=z+1$ the integral becomes

$$I = \int\_{-1}^1F(z)dz=\int\_{-1}^1\dfrac{e^{-1/(z+1)}}{z+1}dz$$

you may use any Gauss-Legendre formula via

$$I=\sum\_{i=0}^nw\_iF(z\_i)$$

For four-point formula when $n=3$, 

$$I=0.5590$$

if you multiply it by $e^{1/2}$, your final value is 0.9216.