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