Approximating $\log x$ with roots The following is a surprisingly good (and simple!) approximation for $\log x+1$ in the region $(-1,1)$:
$$\log (x+1) \approx \frac{x}{\sqrt{x+1}}$$
Three questions:


*

*Is there a good reason why this would be the case?

*How does one go about constructing the "next term"?

*Are the any papers on "generalized Pade approximations" that involve radicals?

 A: Let's rewrite both sides in terms of $y = x + 1$: we get
$$\log y \approx \sqrt{y} - \frac{1}{\sqrt{y}}$$
on, let's say, the interval $\left( \frac{1}{2}, 2 \right)$ (I hesitate to discuss the entire interval $(0, 2)$; it seems to me that the approximation is not all that good near $0$). The RHS should look sort of familiar: let's perform a second substitution $y = e^{2z}$ to get
$$2z \approx e^z - e^{-z} = 2 \sinh z$$
on the interval $\left( - \varepsilon, \varepsilon \right)$ where $\varepsilon = \frac{\log 2}{2} \approx 0.346 \dots$. Of course now we see that the LHS is just the first term in the Taylor series of the RHS, and on a smaller interval than originally. Furthermore, the Taylor coefficients of $2 \sinh z$, unlike the Taylor coefficients of our original functions, decrease quite rapidly. The next term is $\frac{z^3}{3}$, which on this interval is at most 
$$\frac{\varepsilon^3}{3} \approx 0.0138 \dots$$
and this is more or less the size of the error in the approximation between $\log 2$ and $\frac{1}{\sqrt{2}}$ obtained by setting $y = 2$, or equivalently $x = 1$. 
With the further substitution $t = \sinh z$, the RHS is just the first term in the Taylor series of the LHS. To get the "next term" we could look at the rest of the Taylor series of $\sinh^{-1} t$. The next term is $- \frac{t^3}{6}$, which gives
$$z \approx \frac{e^z - e^{-z}}{2} - \frac{(e^z - e^{-z})^3}{48}$$
or
$$\log y \approx \left( \sqrt{y} - \frac{1}{\sqrt{y}} \right) - \frac{1}{24} \left( \sqrt{y} - \frac{1}{\sqrt{y}} \right)^3.$$
I don't know if this is useful for anything. The series to all orders just expresses the identity
$$\log y = 2 \sinh^{-1} \frac{\left( \sqrt{y} - \frac{1}{\sqrt{y}} \right)}{2}.$$
A: The simplest Pade approximant we could build seems to be $$\log(1+x)\approx\frac{x}{1+\frac{x}{2}}$$ and we can notice the similarity of denominators close to $x=0$.
However, the approximation given in the post seems to be significantly better for $x<\frac 12$.
A: A different approach.
The function $f(x)=x-\log(1+x)\sqrt{1+x}$ is continuous and increasing on $[-1,1]$ (I have not proved it, but a graph of $f'$ is sufficiently convincing.) For any $a\in(0,1)$
$$
f(-a)\le f(x)\le f(1)=0.0197419,\quad-a\le x\le1.
$$
Take for instance $a=0.8$ we obtain
$$
-\frac{0.0802375}{\sqrt{1+x}}\le\frac{x}{\sqrt{1+x}}-\log(1+x)\le\frac{0.0197419}{\sqrt{1+x}},\quad -.8\le x\le1.
$$
This shows that $\log(1+x)\sqrt{1+x}$ is a good approximation of $x$.
$$
f(x)=-\frac{x^3}{24}+\dots
$$
is an alternate series. This explains the vey good approximation for $x>0$, and the not so good for $x<0$.
A: I am being rather late, yet still there is some interesting information I can add.
The Padet approximant is of the form $\log(1+x)\approx \frac{x}{1+x/2}$ as noted by other posters. This partially explains why $\frac{x}{\sqrt{1+x}}$ is a good approximation, what it does not explain is why the square-root approximation is better than a supposedly "great" Pade approximation. @nbubis had an idea that it works better because it has pole in the correct spot, but it seems that it is actually a red herring.
Let's take a look on more general Pade $(1,n)$ approximation, it equals $\log(1+x)\approx\frac{x}{1+x/2-x^2/12+x^4/24+...}$.
Now the reason $\frac{x}{\sqrt{1+x}}$ approximation performs better can be explained by $\sqrt{1+x} \approx 1+x/2 -x^2/8$ and noting that $1+x/2-x^2/8$ is closer to the "true value" of the denominator than the first Pade approximant $1+x/2$. 
To see that it is indeed the case consider the approximation 
$\log(1+x)=\frac{x}{(1+5x/6)^{3/5}}$. 
Now, as you can quickly check, $(1+5x/6)^{3/5}$ has the Taylor expansion $\approx 1+x/2 -x^2/12$ which aggrees with first 3 terms of $(1,n)$ Pade approximant. Now if you plot it it will turn out that it is even better than originally suggested $\frac{x}{\sqrt{1+x}}$ despite having pole in the wrong place.  
Regarding method by @QiaochuYuan, you can perform the same "trick" to  get better approximations which will be performing better in the neighbourhood of $x=0$ but worse when $x$ is large, for example
$\log (1+x) \approx \frac{x}{(1+5x/6)^{3/5}} + \frac{x^4}{108 (1+5x/6)^{12/5}}$
But in disguise what you are actually making is finding better approximations to some Pade approximant.
Some of the other approximations you can find in the same way are
$\log(1+x)\approx \frac{x}{\sqrt{1+x+x^2/12}}$ and $\log(1+x)\approx \frac{x}{(1+3x/2+x^2/2)^{1/3}}$ which are good simply beause they coincide with Pade approximant up to the terms of high order. I guess the first among those two is another reason why $\frac{x}{\sqrt{1+x}}$ worked so well.
Short version: It's actually a coincidence, $\sqrt{1+x}$ happen to have Taylor expansion $\sqrt{1+x}\approx 1+x/2-x^2/8$ which coincides with expansion of $x/\log(1+x)\approx 1+x/2 -x^2/12$ up to 2 terms and the third term is not that different to mess up the approximation.  
A: One reason may be that their Taylor series around $x=0$ start the same:
$$
\log (x+1) \approx x-x^2/2+x^3/3+\cdots
$$
$$
 \frac{x}{\sqrt{x+1}} \approx x-x^2/2+3x^3/8+\cdots
$$
So they agree to order 2 for $|x|<1$. They almost agree to order 3 because $1/3 \approx 3/8$ roughly.
However, this is an a posteriori reason. I don't know why this approximation should be good a priori.
