Answer
$$\int \cosh ^{n} x d x=\frac{1}{n} \cosh ^{n-1} x \sinh x+\frac{n-1}{n} \int \cosh ^{n-2} x d x$$
Work Step by Step
Given $$\int \cosh ^{n} x d x $$
Let
\begin{align*}
u&=\cosh^{n-1}x \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ dv=\cosh xdx\\
du&= (n-1)\cosh^{n-2}x \sinh xdx\ \ \ \ \ \ \ \ \ v= \sinh x
\end{align*}
Then
\begin{align*}
\int \cosh ^{n} x d x&= \cosh^{n-1}x \sinh x-(n-1)\int \cosh^{n-2}x \sinh^2 xdx\\
&= \cosh^{n-1}x \sinh x-(n-1)\int \cosh^{n-2}x( \cosh^2 x+1)dx\\
(n )\int \cosh ^{n} x d x&= \cosh^{n-1}x \sinh x-(n-1)\int \cosh^{n-2}x dx\\
\end{align*}
Hence $$\int \cosh ^{n} x d x=\frac{1}{n} \cosh ^{n-1} x \sinh x+\frac{n-1}{n} \int \cosh ^{n-2} x d x$$
