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

$$\frac{1}{3} \ln |3 \sinh x+4|+C $$

Given $$ \int \frac{\cosh x}{3 \sinh x+4} d x$$ Let $$ u=3 \sinh x+4 \ \ \ \to du = 3\cosh x dx$$ Then \begin{aligned} \int \frac{\cosh x}{3 \sinh x+4} d x &=\frac{1}{3} \int \frac{1}{u} d u \\ &=\frac{1}{3} \ln |u|+C \\ &=\frac{1}{3} \ln |3 \sinh x+4|+C \end{aligned}