Single Variable Calculus, Chapter 7, 7.3-2, Section 7.3-2, Problem 82

Find the integral $\displaystyle \int \frac{e^{\frac{1}{x}}}{x^2} dx$

If we let $\displaystyle \frac{1}{x}$, then $\displaystyle du = \frac{-1}{x^2} dx$, so $\displaystyle \frac{1}{x^2} dx = -du$. Thus,


$
\begin{equation}
\begin{aligned}

\int \frac{e^{\frac{1}{x}}}{x^2} dx =& \int e^{\frac{1}{x}} \frac{1}{x^2} dx
\\
\\
\int \frac{e^{\frac{1}{x}}}{x^2} dx =& \int e^u \cdot -du
\\
\\
\int \frac{e^{\frac{1}{x}}}{x^2} dx =& - \int e^u du
\\
\\
\int \frac{e^{\frac{1}{x}}}{x^2} dx =& -e^u + C
\\
\\
\int \frac{e^{\frac{1}{x}}}{x^2} dx =& -e^{\frac{1}{x}} + C


\end{aligned}
\end{equation}
$