Chapter 8 - Techniques of Integration - 8.4 Integrals Involving Hyperbolic and Inverse Hyperbolic Functions - Exercises - Page 415: 6

$$-\frac{1}{3} \operatorname{sech}(3 x)+C$$

Given $$\int \operatorname{sech}^{2}(1-2 x) d x$$ Let $$ u=3x \ \ \ \to du =3dx$$ Then \begin{aligned} \int \tanh (3 x) \operatorname{sech}(3 x) d x &=\frac{1}{3} \int \tanh (u) \operatorname{sech}(u) d x \\ &=-\frac{1}{3} \operatorname{sech}(u)+C\\ &=-\frac{1}{3} \operatorname{sech}(3 x)+C \end{aligned}