# find $\int{artanh(x)} dx$

I've been asked to find $\int{artanh(x)} dx$

The first thing I did was I said:

let $y = artanh(x)$

$\therefore \dfrac{dy}{dx} = \dfrac{1}{1-x^2}$

$\therefore dx = (1-x^2)dy$

Also, from $y = artanh(x)$ I know that:

$tanh(y) = x$

$\therefore 1-x^2 = 1-tanh^2(y)$

From this is can therefore say:

$\int{artanh(x)} dx = \int{y(1-tanh^2y)}dy$

$= \int{y- y(tanh^2y)}dy$

$= \int{y}dy - \int{y(tanh^2y)}dy$

$= \frac{1}{2}y^2 - \int{y(tanh^2y)}dy$

To solve $\int{y(tanh^2y)}dy$ I tried integrating by parts, so I said:

let $u = y$ and let $\dfrac{dv}{dy} = tanh^2y$

$\therefore \dfrac{du}{dy} = 1$ and $v = \int{1 - sech^2y}dy$

$\therefore v = y - tanhy$

If I substitute these values into the equation:

$uv - \int{v}{\dfrac{du}{dx}}$

I get:

$\int{ytanh^2y}dy = ytanh^2y - \int{y - tanhy}dy$

I can therefore say that:

$\int{artanh(x)} dx = \frac{1}{2}y^2 - ytanh^2y + \int{y - tanhy}dy$

$= \frac{1}{2}y^2 -ytanh^2y + \frac{1}{2}y^2 -lncoshy + C$

$= y^2 -ytanh^2y -lncoshy + C$

But $y = artanhx$

$\therefore y^2 = artanh^2x$

and $tanh^2y = (tanh(artanhx))^2 = x^2$

and $coshy = cosh(artanhx)$

To simplify this part I used substitution:

let $u = artanhx$

I'm therefore trying to find $cosh(u)$

from $u = artanhx$ I know that:

$tanhu = x$

$\therefore tanh^2u = x^2$

$\therefore 1- sech^2u = x^2$

$\therefore cosh^2u = \dfrac{1}{1-x^2}$

$\therefore coshu = \dfrac{1}{\sqrt{1-x^2}}$

From this I can therefore say that:

$\int{artanh(x)} dx = artanh^2x - x^2(artanhx) - ln(\dfrac{1}{\sqrt{1-x^2}}) + C$

$\therefore \int{artanh(x)} dx = artanhx(artanhx - x^2) + \frac{1}{2}ln(1-x^2) + C$

$\int{artanh(x)} dx = x(artanhx) + \frac{1}{2}ln(1-x^2) + C$

I can't see where I've gone wrong. I understand I've probably made this more difficult than it needs to be and therefore probably made a silly mistake somewhere. I've probably missed something that makes the whole question a lot simpler but I can't see what that would be.

Thank you :)

Why don't you integrate by parts using $u=\tanh ^{-1}(x)$ and $v'=dx$.