# Math Help - Integration of inverse hyperbolic functions

1. ## Integration of inverse hyperbolic functions

Prove that:
$\displaystyle \int^1_{\frac{4}{5}} arsech xdx=2\arctan 2-\frac{\pi}{2}-\frac{4}{5}\ln 2$

I found $arsech x = \ln{\frac{1+\sqrt{1+x^2}}{x}}$
But When I tried to integrate it, it got very messy as well as not being the answer. I've got no idea how else to integrate, so any help will be very much appreciated.
Thanks!

2. Use integration by parts with $u = \textrm{arsech}\,{x}$ and $dv = 1$.

3. $u=arsech x \rightarrow \frac{du}{dx}=-\frac{1}{x\sqrt{1-x^2}}$
$\frac{dv}{dx}=1 \rightarrow v=x$

Then the integral
$x arsech x -\int -\frac{1}{\sqrt{1-x^2}}$
$=x arsech x + \arcsin x$

Filling in the values, I get:
$\arctan \frac{4}{3}-\frac{\pi}{2}-\frac{4}{5}\ln 2$
Which is correct except for
$\arctan \frac{4}{3}$