Chapter 5 - Review - Exercises - Page 430: 36

$\int sinh (1+4x)~dx = \frac{1}{4}cosh (1+4x)+C$

$\int sinh (1+4x)~dx$ Let $u = 1+4x$ $\frac{du}{dx} = 4$ $dx = \frac{du}{4}$ $\int~sinh ~u~\frac{du}{4}$ $= \int~\frac{1}{4}~sinh ~u~du$ $= \frac{1}{4}cosh ~u+C$ $= \frac{1}{4}cosh (1+4x)+C$