A Fun Way of Finding the Artanh Integral

When hoping for the best works out

Last updated 2025 June 10

This is the artanh integral, which shows up pretty often, especially when integrating rational functions:

∫11−x2 dxint rac{1}{1-x^2} dx

The canonical method is to use partial fraction decomposition:

=∫1(1−x)(1+x) dx=∫1(1+x)(1−x) dx=12∫11+x+11−x dxegin{align*} &= int rac{1}{(1-x)(1+x)} dx \ &= int rac{1}{(1+x)(1-x)} dx \ &= rac{1}{2} int rac{1}{1+x} + rac{1}{1-x} dx end{align*}

This then integrates easily via 2 straight-up layer cakes, into a form that happens to collapse to tanh⁡−1(x)\tanh^{-1}(x).

=12[ln⁡(1+x)−ln⁡(1−x)]=12ln⁡(1+x1−x)=tanh⁡−1x+cegin{align*} &= rac{1}{2} left[ ln(1+x) - ln(1-x) ight] \ &= rac{1}{2}lnleft( rac{1+x}{1-x} ight) \ &= anh^{-1}{x} + c end{align*}

I’ve found a different route1 to this result, which came to me while thinking about polynomial division.

Pulling the Power Move

My thought process was this – you’ve got these factors in the denominator, so it’d be great if you could just cancel them with stuff in the numerator.

∫1(1−x)(1+x) dxint rac{1}{(1-x)(1+x)} dx

Well, the easiest way to do that would be to get a (1+x)(1+x) in the numerator – so we’ll add (+x−x)(+x -x).

=∫1+x−x(1−x)(1+x) dx= int rac{1+x-x}{(1-x)(1+x)} dx

Now splitting spits out a nicely integratable layer cake on the left, but the right is still a bit more problematic.

=∫1+x(1−x)(1+x)+−x(1−x)(1+x) dx=∫11−x+−x(1−x)(1+x) dxegin{align*} &= int rac{1+x}{(1-x)(1+x)} + rac{-x}{(1-x)(1+x)} dx \ &= int rac{1}{1-x} + rac{-x}{(1-x)(1+x)} dx end{align*}

What we want is an extra 11 to get (1−x)(1-x). So… let’s add (+1−1)(+1-1).

=∫11−x+−x+1−1(1−x)(1+x) dx=∫11−x+1−x(1−x)(1+x)+−1(1−x)(1+x) dx=∫11−x+11+x dx−∫1(1−x)(1+x) dxegin{align*} &= int rac{1}{1-x} + rac{-x+1-1}{(1-x)(1+x)} dx \ &= int rac{1}{1-x} + rac{1-x}{(1-x)(1+x)} + rac{-1}{(1-x)(1+x)} dx \ &= int rac{1}{1-x} + rac{1}{1+x} dx - int rac{1}{(1-x)(1+x)} dx end{align*}

What’d’y’know, that’s our original integral! What an incredible duplication we’ve got on our hands. So, moving to the other side, we have

∫1(1−x)(1+x) dx=∫11−x+11+x dx−∫1(1−x)(1+x) dx2∫1(1−x)(1+x) dx=∫11−x+11+x dx2∫1(1−x)(1+x) dx=−ln⁡(1−x)+ln⁡(1+x)∫1(1−x)(1+x) dx=12ln⁡(1+x1−x)+cegin{align*} extcolor{#4d9dcd}{int rac{1}{(1-x)(1+x)} dx} &= int rac{1}{1-x} + rac{1}{1+x} dx - extcolor{#4d9dcd}{int rac{1}{(1-x)(1+x)} dx} \ extcolor{#4d9dcd}{2 int rac{1}{(1-x)(1+x)} dx} &= int rac{1}{1-x} + rac{1}{1+x} dx \ 2int rac{1}{(1-x)(1+x)} dx &= -ln(1-x) + ln(1+x) \ int rac{1}{(1-x)(1+x)} dx &= rac{1}{2}lnleft( rac{1+x}{1-x} ight) + c end{align*}

The Original Power Move

That’s a nice way of seeing how you could’ve derived it through problem-solving. I actually got there more through instinct/intuition – after staring at my phone for a couple seconds and pondering the possibility, I realised we could get the extra 11 we needed by multiplying by 22 at the start.

 ∫1(1−x)(1+x) dx= 12∫2(1−x)(1+x) dxegin{align*} & int rac{1}{(1-x)(1+x)} dx \ =& rac{1}{2} int rac{2}{(1-x)(1+x)} dx end{align*}

Then you pull the power move, and the split is now strategical:

=12∫2+x−x(1−x)(1+x) dx=12∫1+x(1−x)(1+x)+1+x(1−x)(1+x) dx=12∫11−x+11+x dxegin{align*} &= rac{1}{2} int rac{2+x-x}{(1-x)(1+x)} dx \ &= rac{1}{2} int rac{1+x}{(1-x)(1+x)} + rac{1+x}{(1-x)(1+x)} dx \ &= rac{1}{2} int rac{1}{1-x} + rac{1}{1+x} dx end{align*}

Just works ;)

Does It Generalise?

Of course if you know the artanh integral it’s a bit of a waste of time to re-derive this relationship every time. But just for enrichment, let’s see if this method generalises.

∫1k2−x2 dxint rac{1}{k^2 - x^2} dx

How should I put this… yeah, ofc it does lmao. Only needed to stare at the problem for like, 4 seconds. Why did I think it wouldn’t :confused:

=∫1(k−x)(k+x) dx=12k∫2k(k−x)(k+x) dx=12k∫2k+x−x(k−x)(k+x) dx=12k∫k+x(k−x)(k+x)+k−x(k−x)(k+x) dx=12k∫1k−x+1k+x dxegin{align*} &= int rac{1}{(k-x)(k+x)} dx \ &= rac{1}{2k} int rac{2k}{(k-x)(k+x)} dx \ &= rac{1}{2k} int rac{2k+x-x}{(k-x)(k+x)} dx \ &= rac{1}{2k} int rac{k+x}{(k-x)(k+x)} + rac{k-x}{(k-x)(k+x)} dx \ &= rac{1}{2k} int rac{1}{k-x} + rac{1}{k+x} dx end{align*}

Love to see it!

  1. *visually different, algebraically it’s all the same lol↩