I am completely done with this problem. I transformed $x^2+1$ to $(x+i)(x-i)$, also used standard form of $n^{th}$ derivative of $\arctan(x)$, again gone negative binomial expansion but nothing is working. If anyone can make it I will be very glad. The question is:
Show that the $n^{th}$ derivative of $1/(x^2+1)$ is equal to $$ \frac{(-1)^n \cdot n!}{(x^2+1)^{n+1}} \cdot \left[(n+1)x^n - \,{}^{n+1}C_3 \,x^{n-2} +\,{}^{n+1}C_5\,x^{n-4} - \cdots \right] $$
The partial fraction expansion is $$f(x)=\frac1{x^2+1}=\frac{1}{2i}\left(\frac1{x-i}-\frac1{x+i}\right).$$ Therefore \begin{align} f^{(n)}(x)&=(-1)^n\frac{n!}{2i} \left(\frac1{(x-i)^{n+1}}-\frac1{(x+i)^{n+1}}\right)\\ &=(-1)^n\frac{n!}{2i(x^2+1)^{n+1}} \left((x+i)^{n+1}-(x-i)^{n+1}\right) \end{align} and you get your formula by expanding the $(x\pm i)^{n+1}$ via the Binomial Theorem.