Answer
$$\frac{1}{4} \cosh ^{3} x \sinh x+\frac{3}{8} \cosh x \sinh x+\frac{3}{8} x+C$$
Work Step by Step
Given $$\int \cosh ^{4} x d x$$
Use
$$\int \cosh ^{n} x d x=\frac{1}{n} \cosh ^{n-1} x \sinh x+\frac{n-1}{n} \int \cosh ^{n-2} x d x$$
Then for $n=4$
\begin{aligned}
\int \cosh ^{4} x d x &=\frac{1}{4} \cosh ^{4-1} x \sinh x+\frac{4-1}{4} \int \cosh ^{4-2} x d x \\
&=\frac{1}{4} \cosh ^{3} x \sinh x+\frac{3}{4} \int \cosh ^{2} x d x \\
&=\frac{1}{4} \cosh ^{3} x \sinh x+\frac{3}{4}\left[\frac{1}{2} \cosh ^{2-1} x \sinh x+\frac{2-1}{2} \int \cosh ^{2-2} x d x\right] \\
&=\frac{1}{4} \cosh ^{3} x \sinh x+\frac{3}{4}\left[\frac{1}{2} \cosh x \sinh x+\frac{1}{2} \int 1 \, d x\right] \\
&=\frac{1}{4} \cosh ^{3} x \sinh x+\frac{3}{8} \cosh x \sinh x+\frac{3}{8} x+C
\end{aligned}
