Answer
$$-e^{{}^{1}{\mskip -5mu/\mskip -3mu}_{t}}+C$$
Work Step by Step
$\color{blue}{\text{Rewrite}}$
$$\int \dfrac{e^{{}^{1}{\mskip -5mu/\mskip -3mu}_{t}}}{t^2}dt=\int e^{{}^{1}{\mskip -5mu/\mskip -3mu}_{t}}\cdot \dfrac{1}{t^2}dt$$
$\color{blue}{\text{Let } u=\dfrac{1}{t}\Rightarrow \quad du=-\dfrac{1}{t^2}dt\Rightarrow \quad -du=\dfrac{1}{t^2}dt}$
$\color{blue}{\text{Make the substitution}}$
$$ \int \underbrace{e^{{}^{1}{\mskip -5mu/\mskip -3mu}_{t}}}_{\color{red}{e^{u}}}\cdot \underbrace{\dfrac{1}{t^2}dt}_{\color{red}{-du}} =\int e^u (-du)=-\int e^u du$$
$\color{blue}{\text{Integrate}}$
$$-\int e^u du=-e^u$$
$\color{blue}{\text{Back-Substitute }\text{ } u=\dfrac{1}{t}}$
$$ \boxed{\int \dfrac{e^{{}^{1}{\mskip -5mu/\mskip -3mu}_{t}}}{t^2}dt=-e^{{}^{1}{\mskip -5mu/\mskip -3mu}_{t}}}$$
