Chapter 8 - Techniques of Integration - 8.4 Integrals Involving Hyperbolic and Inverse Hyperbolic Functions - Exercises - Page 415: 7

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

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