$n$-th derivative of $\log(1+x)/x$ What is the $n$-th derivative of  $$\frac{\log(1+x)}{x}$$
Now I have seen some analytical methods of getting $n$-th derivative of nicer looking functions such as the $n$-th derivative of  $$ 1\over 1+x $$
However, this is not the case for $$\frac{\log(1+x)}{x}$$
 A: Since
$$ \frac{\log(1+x)}{x}=\int_{0}^{1}\frac{dt}{1+xt} \tag{1} $$
we have:
$$ \frac{d^n}{dx^n}\frac{\log(1+x)}{x}=(-1)^n n! \int_{0}^{1}\frac{t^n}{(1+tx)^{n+1}}\,dt \tag{2}$$
and:
$$ \int_{0}^{1}\frac{t^n}{(1+tx)^{n+1}}\,dt = \frac{1}{x^n}\int_{0}^{1}\frac{((1+tx)-1)^n}{(1+tx)^{n+1}}\,dt \tag{3}$$
can be easily computed through the binomial theorem.
A: Another form is to consider the $n^{th}$ derivative of a product of functions. This leads to the form
$$D^{n}\{f(x) \, g(x)\} = \sum_{r=0}^{n} \binom{n}{r} \, D^{r}\{f(x)\} \, D^{n-r}\{g(x)\}.$$
Let $g(x) = 1/x$ and $f(x) = \ln(1+x)$ for which
\begin{align}
g^{(n)}(x) &= \frac{(-1)^{n} \, n!}{x^{n+1}} \\
f^{(0)}(x) &= \ln(1+x) \\
f^{(n)}(x) &= \frac{(-1)^{n-1} \, (n-1)!}{(1+x)^{n}}, n\geq 1. 
\end{align}
Now,
\begin{align}
D^{(n)}\left\{ \frac{\ln(1+x)}{x} \right\} &= D^{(n)}\left\{\frac{1}{x}\right\} + \sum_{r=1}^{n} \binom{n}{r} \, D^{(r)}\{\ln(1+x)\} \, D^{(n-r)}\left\{ \frac{1}{x} \right\} \\
&= \frac{(-1)^{n} \, n!}{x^{n+1}} \, \left[ \ln(1+x) - \sum_{r=1}^{n} \frac{1}{r} \, \left(\frac{x}{1+x}\right)^{r} \right].
\end{align}
