I know the Taylor series of $\arctan x$ at $x=0$.
But I have been stuck when $x=1$.
I computed $f'(1)=1/2$, $f''(1)=-1/2$, $f'''(1)=1/2$, $f''''(1)=0$, $f'''''(1)=-3$.
But.. I can't find any regularity.
What should I do to write regular Taylor series form..?
Start from expanding the derivative of $\arctan(x)$, which is $\frac{1}{1+x^2}$.
To expand it around $x=1$ write it as
$$\begin{align}\frac{1}{1+x^2}&=\frac{i/2}{i+x}+\frac{-i/2}{-i+x}\\ &=\frac{i/2}{i+1+x-1}+\frac{-i/2}{-i+1+x-1}\\ &=\frac{i/2}{(i+1)}\frac{1}{1+(x-1)/(i+1)}+\frac{-i/2}{-i+1}\frac{1}{1+(x-1)/(1-i)}\\ &=\frac{i/2}{(i+1)}\sum_{n=0}^{\infty}\frac{(-1)^n(x-1)^n}{(1+i)^n}+\frac{-i/2}{-i+1}\sum_{n=0}^{\infty}\frac{(-1)^n(x-1)^n}{(1-i)}\\ &=\sum_{n=0}^{\infty}\left(\frac{(-1)^ni/2}{(i+1)^{n+1}}+\frac{(-1)^n(-i/2)}{(-i+1)^{n+1}}\right)(x-1)^{n}\end{align}$$
Now you can integrate term-by-term and add the constant $\arctan(1)$ to get
$$\arctan(x)=\arctan(1)+\sum_{n=0}^{\infty}\frac{\left(\frac{(-1)^ni/2}{(i+1)^{n+1}}+\frac{(-1)^n(-i/2)}{(-i+1)^{n+1}}\right)}{n+1}(x-1)^{n+1}$$
Note that all coefficients are real numbers. You can simplify them further if you like.