I don't understand this property of arctan(x)'

Yes, I would like to see your derivation of B'. I get nothing like that!

 

Wolfram solution

[itex]\frac{d}{dx}tan^{-1}(x-\sqrt{x^2+1}) = \frac{1-\frac{x}{\sqrt{x^2+1}}}{(x-\sqrt{x^2+1})^2+1}[/itex]

 

That's what I came out with!

 

I plugged this into mathematica and it indeed simplifies to [tex]\frac{1}{2+2x^2}[/tex]. Which would suggest that [itex]\arctan(x)=2 \arctan(x-\sqrt{x^2+1})+C[/itex]. Therefore [itex] C=\arctan(x)-2\arctan(x-\sqrt{x^2+1})=\frac{\pi}{4}[/itex].

Picture:
http://img32.imageshack.us/img32/3028/arctan.jpg [Broken]

I am not sure how to find C yet, mathematica wasn't able to find it and I am not in the mood to try and symplify it currently. However all the numerical values I tried yielded pi/4.

 

One way is to set

[tex]
z=\tan^{-1} \left(x - \sqrt{1 + x^2} \right),
[/tex]

so that

[tex]\tan z = x - \sqrt{1 + x^2}[/tex]

Solving for [itex]x[/itex] then leads directly to your result.

 

That's a clever way to solve it George.

Mr_garlic, your first result was correct. After simplification that ugly looking expression will reduce to the nice expression you found in your first post.