Vtyjkyuk

I'm trying to work out how to find the indefinite integral of $operatornamearccosh(x + 1)$

I have been using integration by parts to get it down to $$xoperatornamearccosh(x + 1) -
int fracxsqrtleft((x+1)^2 - 1 right) , dx$$

What I'm unsure of is how to integrate the last part

Help would be much appreciated!

First do the substitution $u=x+1$; then
$$
intarccos(x+1),dx=
intarccos u,du=
uarccos u+intfracusqrt1-u^2,du
=uarccos u-sqrt1-u^2+c
$$
Back substitute and you're done.

For $operatornamearcosh$ it's essentially the same, but using that the derivative of $f(t)=operatornamearcosht$ is
$$
f'(t)=frac1sqrtt^2-1
$$
Therefore
$$
intoperatornamearcosh(x+1),dx=
intoperatornamearcosh u,du=
uoperatornamearcosh u-intfracusqrtu^2-1,du
=uarccos u-sqrtu^2-1+c
$$

The function $g(t)=sqrt1-t^2$ has derivative
$$
g'(t)=-fractsqrt1-t^2
$$
so the final integral in the first part is immediate. Similarly for the second part.

Another quite intuitive way is just to use the substitution $u=cosh^-1(x+1)$ so that $x=cosh u -1$ and $dx=sinh u;du$. Now
$$intcosh^-1(x+1)dx = int usinh u;du = ucosh u-int cosh u ; du = ucosh u-sinh u $$
so your solution is
$$ (x+1)left(cosh^-1(x+1)right)-sqrt(x+1)^2-1. $$