Chapter 8 - Techniques of Integration - 8.4 Integrals Involving Hyperbolic and Inverse Hyperbolic Functions - Exercises - Page 415: 5

$$\frac{-1}{2}\tanh (1-2x)+C$$

Given $$\int \operatorname{sech}^{2}(1-2 x) d x$$ Let $$ u=1-2x \ \ \ \to du =-2dx$$ Then \begin{aligned} \int \operatorname{sech}^{2}(1-2 x) d x&=-\frac{1}{2} \int \operatorname{sech}^{2}(u) d u\\ &=\frac{-1}{2}\tanh u+C\\ &=\frac{-1}{2}\tanh (1-2x)+C \end{aligned}