Chapter 7 - Exponential Functions - 7.9 Hyperbolic Functions - Exercises - Page 384: 42

$$\int \frac{\cosh x}{3\sinh x +4} dx =\frac{1}{3} \ln (3\sinh x +4)+c $$

Let $ u=3\sinh x +4$, then $ du=3\cosh x dx $ and hence $$\int \frac{\cosh x}{3\sinh x +4} dx = \frac{1}{3}\int \frac{du}{u}=\frac{1}{3} \ln u+c\\ =\frac{1}{3} \ln (3\sinh x +4)+c $$