A Fun Way of Finding the Artanh Integral

When hoping for the best works out

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↩