What is the $n$ th derivative of $\ln(x)/(1+x^2)\:?$ I'm into something but I came across the problem of finding a closed form for
$$\left( \frac{\ln x}{1+x^2} \right)^{(n)}$$
where the little $(n)$ denotes the $n$ th derivative of the function. After looking in some softwares, I believe the closed form is a rational function with $\ln x$ as a factor.
Thank for any help.
Edit: I found a way to change the problem above to this one:
$$ \left( \frac{\partial }{\partial x}\frac{(1-x)}{(1+x^2)(1-(1-x)y)}\right)^{(n)}$$
maybe this is easier?
EDIT $2$: I said closed formula but forgot to mention something easy to handle, and a summation like $\sum_{k=0}^n \cdots$ is hard, because after finding the derivative I'll need to find it's maximum...
A good example of what I mean Finding the $2n+1$ th derivative of $\frac{y^{2n+1}xy}{1-x^2y^2}$ with respect to $x$.
I'm trying to establish the irrationality of a constant using Beuker's integrals, and using this approach you'll eventually need to find the $n$ th derivative of a function. In this case it can be either the first function or the second showed here. The $y$ in the second came from a substitution in the Beuker's integral.
 A: If this can help,
$$(x^2+1)f(x)=\ln(x)$$
$$2xf(x)+(x^2+1)f'(x)=\frac1x$$
$$2f(x)+4xf'(x)+(x^2+1)f''(x)=-\frac1{x^2}$$
and more generally
$$(k+1)(k+2)f^{(k)}(x)+2(k+2)xf^{(k+1)}(x)+(x^2+1)f^{(k+2)}(x)=\frac{(-1)^{k+1}}{x^{k+2}}.$$
A: Use the Leibniz rule
$\def\d{\frac{d^n}{dx^n}}$
$\def\cc#1{\left[#1 - \mathit{c.\!c.}\right]}$
$\def\lnn#1{\underline{\ln}_{#1}}$
$\def\Im{\operatorname{Im}}$
$\def\Re{\operatorname{Re}}$
$\def\im#1{\Im\left( #1 \right)}$
$$\begin{align}
(fg)^{(n)}
&=\sum_{k=0}^n \binom n k f^{(n-k)} g^{(k)}\\
&=f^{(n)}g + \sum_{k=1}^n \binom n k f^{(n-k)} g^{(k)}
\end{align}$$
for the product. For $n\geqslant1$, the $n$-th derivative of $g=\ln$ is:
$$
\d\ln x=(-1)^{n+1}\frac{(n-1)!}{x^n}
$$
From that we also get the $n$-th derivative of $f(x)=(1+x^2)^{-1}$ as
$$
\d\frac1{x^2+1} = (-1)^n n! \Im{\frac{1}{(x-i)^{n+1}}}
$$
by noticing that
$$
\frac1{x^2+1}
= \frac{i}{2}\left(\frac{1}{x+i} - \frac{1}{x-i}\right)
= \Im{\frac1{x-i}}
$$
Using the imaginary part only works for real $x$, and in the remainder I'll use $\Im$ assuming $x\in\mathbb R$.  For complex $x$ it's the same route, just more tedious to write down.  Plugging everything into Leibniz' rule, the binomial coefficients cancel out almost entirely and we get
$$
\d\frac{\ln x}{x^2+1}
=(-1)^n n! \im{\frac{1}{(x-i)^{n+1}}
\left(
\ln x - \sum_{k=1}^n \frac 1k \left(1-i/x \right)^k
\right)}
$$

Writing the sum as incomplete lower Natural logarithm
  $$
\lnn{n} x = -\sum_{k=1}^n \frac 1k (1-x)^k
$$
  we get for $x\in\mathbb R$:
  $$
\d\frac{\ln x}{x^2+1}
=(-1)^n n! \Im \frac{\ln x + \lnn{n}(i/x)}{(x-i)^{n+1}}
$$

Moreover, this gives an asymptotic formula for $|x|<1$ due to $\lnn{\infty}=\ln$:
$$
\d\frac{\ln x}{x^2+1} \approx
\Re \frac{(-1)^n n! \pi}{2(x-i)^{n+1}} \qquad\text{for large }n
$$
A: Find the nth derivative of $ln(x)$ and $(1+x^2)^{(-1)}$ and then use Liebnitz's rule
$(fg)^{(n)}
= \sum_{k=0}^n \binom{n}{k}f^{(k)}g^{(n-k)}$.
