Chapter 7 - Exponential Functions - 7.9 Hyperbolic Functions - Exercises - Page 384: 40

$$\int \tanh 3x \, sech\, 3x dx \ =-\frac{1}{3}sech \, 3x+c.$$

Let $ u=3x $, then $ du=3dx $ and hence $$\int \tanh 3x \, sech\, 3x dx =\frac{1}{3}\int \tanh u \, sech\,u dxdu =-\frac{1}{3}sech \, u+c\\ =-\frac{1}{3}sech \, 3x+c.$$