Can this standard calculus result be explained "intuitively" Recently I stumbled upon someone who said he wanted to understand why
$\arctan x = \int\dfrac{dx}{1+x^2}$
At first I was confused. This is an easy result in any integral calculus course. But then he explained that although he understood the proof, he wanted to understand it "intuitively". He wanted to see why it was in terms of arclength and addition and subtraction.
My question is: Is there an "intuitive" way to explain this identity?
 A: Stereographic projection offers a geometric interpretation of this formula. In fact the factor
$(1+x^2)^{-1}$ is the metric of the circle in the stereographic projection coordinate. The construction is as follows:
Take a circle of radius 1/2, call the angle of (the radius vector of) a general 
point on the circle with respect to the radius to the south pole $2\theta$.
Connect the point with the north pole and call the distance of the point of intersection of the extension
of this chord with the tangent at the south pole $x$. One can parameterize the points on the circle
by the coordinate $x$ which is called the stereographic projection coordinate. It is easy to see that $ x = tan(\theta)$.
The length of an arc of the circle between $\theta_1$ and$\theta_2$ (having the stereographic projection coordinates $x_1$ and $x_2$) is given by:
$s = 2 \int_{\theta_1}^{\theta_2} d\theta = 2\int_{x_1}^{x_2} \frac{dx}{1+x^2}$ 
This construction has further nice properties


*

*The group of plane rotations $SO(2)$ acts on $x$ by a Möbius transformation.

*This construction generalizes to spheres in higher dimensions.
A: 
A: Put $\arctan x=:\alpha$, let $n$ be large, choose points $t_k:=\tan(k\alpha/n)$  $\ (0\leq k\leq n)$ and put  $\tau_k:=\sqrt{t_k t_{k-1}}$. Then
$$\sum_{k=1}^n  {t_k - t_{k-1} \over 1 + \tau_k^2}=\sum_{k=1}^n  {t_k - t_{k-1} \over 1 + t_k t_{k-1}}= n \tan(\alpha/n).$$
Here the left side is a Riemann sum for the integral $$\int_0^{\tan\alpha}{dt \over 1+t^2}=\int_0^x{dt \over 1+t^2},$$ and the right side has limit $\alpha=\arctan x$ when $n\to\infty$.
A: Since $D \tan(x) = 1 + \tan(x)^2$ and $\tan(\arctan(x)) = x$ we see that:
$$\begin{align}
\tan(\arctan(x)) &= x \\
(1 + \tan(\arctan(x))^2 \cdot D \arctan(x) &= 1 \\
D \arctan(x) &= \frac{1}{(1 + x)^2} \\
\end{align}$$
So by the fundamental theorem of calculus we recover the integral.
A: The geometric picture is as follows.  Let $O = (0, 0), A = (1, 0), X = (1, x)$.  Then $\arctan x = \angle AOX$.  We want to understand why $\angle AOY \approx \arctan x + \frac{h}{1 + x^2}$ where $Y = (1, x + h)$ and $h$ is small; equivalently, we want to understand why $\angle XOY \approx \frac{h}{1 + x^2}$.  Since this angle is small, we equivalently want to understand why $\sin \angle XOY \approx \frac{h}{1 + x^2}$.
Now $\triangle XOY$ evidently has area $\frac{h}{2}$.  On the other hand, it has area $\frac{1}{2} |OX| |OY| \sin \angle XOY$ where $|OX| = \sqrt{1 + x^2}$ and $|OY| \approx |OX|$.  The result follows.
(The derivative follows, anyway.  The integral follows by dividing up $AX$ into little pieces and drawing a bunch of lines to $O$, then summing up all of the contributions.)
A: The real answer I think involves complex numbers. Factor $1+x^2 = (i-x)(-i-x)$, decompose in partial fractions, get an expression involving logarithms that ends up being $\arctan x$ as described in http://en.wikipedia.org/wiki/Inverse_trigonometric_functions#Logarithmic_forms.
A: This was actually my question originally.  To put the demand for intuitiveness in starker terms:  what does the arc of a circle have to do with the area of a square and the operation of reciprocation?
If someone can show me a geometric figure with area $1 \over 1 + x^2$ that might be a good start.
Or, they can explain how this sequence


*

*Times a number by itself;

*Add one to it;

*Flip that result


leads to "a little bit of" circular arc.
