Integral of $\arccos(x + 1)$

Question

I'm trying to work out how to find the indefinite integral of $\operatorname{arccosh}(x + 1)$

I have been using integration by parts to get it down to $$x\operatorname{arccosh}(x + 1) - \int \frac{x}{\sqrt{\left({(x+1)^2} - 1 \right)}} \, dx$$

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

Help would be much appreciated!

Answer 1

First do the substitution $u=x+1$; then $$ \int\arccos(x+1)\,dx= \int\arccos u\,du= u\arccos u+\int\frac{u}{\sqrt{1-u^2}}\,du =u\arccos u-\sqrt{1-u^2}+c $$ Back substitute and you're done.

For $\operatorname{arcosh}$ it's essentially the same, but using that the derivative of $f(t)=\operatorname{arcosh}t$ is $$ f'(t)=\frac{1}{\sqrt{t^2-1}} $$ Therefore $$ \int\operatorname{arcosh}(x+1)\,dx= \int\operatorname{arcosh} u\,du= u\operatorname{arcosh} u-\int\frac{u}{\sqrt{u^2-1}}\,du =u\arccos u-\sqrt{u^2-1}+c $$

The function $g(t)=\sqrt{1-t^2}$ has derivative $$ g'(t)=-\frac{t}{\sqrt{1-t^2}} $$ so the final integral in the first part is immediate. Similarly for the second part.

Answer 2

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 $$\int\cosh^{-1}(x+1)dx = \int u\sinh u\;du = u\cosh u-\int \cosh u \; du = u\cosh u-\sinh u $$ so your solution is $$ (x+1)\left(\cosh^{-1}(x+1)\right)-\sqrt{(x+1)^2-1}. $$