Chapter 5 - Section 5.5 - Indefinite Integrals and the Substitution Method - Exercises - Page 330: 54

$$\int \frac{1}{x^2}e^{1/x}\sec(1+e^{1/x})\tan(1+e^{1/x})dx=-\sec(1+e^{1/x})+C$$

$$A=\int \frac{1}{x^2}e^{1/x}\sec(1+e^{1/x})\tan(1+e^{1/x})dx$$ We set $u=1+e^{1/x}$ Then $$du=e^{1/x}\Big(\frac{1}{x}\Big)'dx=-\frac{1}{x^2}e^{1/x}dx$$ $$\frac{1}{x^2}e^{1/x}dx=-du$$ Therefore, $$A=-\int\sec u\tan udu$$ $$A=-\sec u+C$$ $$A=-\sec(1+e^{1/x})+C$$