Approximating $\log x$ with roots

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

$$\log y = 2 \sinh^{-1} \frac{\left( \sqrt{y} - \frac{1}{\sqrt{y}} \right)}{2}.$$