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?
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

$$\log y = 2 \sinh^{-1} \frac{\left( \sqrt{y} - \frac{1}{\sqrt{y}} \right)}{2}.$$
$$\log(x+1)=\lim_{n\to\infty}n(\sqrt[n]{x+1}-1).$$

In the case $n=2$,
$$2(\sqrt{x+1}-1)=2\frac{x}{\sqrt{x+1}+1}\approx\frac x{\sqrt{x+1}}$$
for small $x$.

The approximation works better as it has a vertical asymptote at $x=-1$.