Find the Taylor series of $f(x)=\arctan (x) $ around $c=1$ Find the Taylor series of $f(x)=\arctan (x) $ around $c=1$. For which $x$ does it converge to  $f(x)$?
This is what I have been able to do so far
$$f(x)=\arctan(x)=\int \frac{1}{1+x^2}\, dx=\int \frac{1}{1+x^2+1-1}\,dx$$
Any hints please???
 A: Hint: The $n$-th degree Taylor polynomial about $c$ is:
$$T_n(x) = f(c) + \frac{f^{\prime}(c)}{1!} (x-c) + \frac{f^{\prime\prime}(c)}{2!} (x-c)^2 + \cdots + \frac{f^{(n)}(x)}{n!} (x-c)^n  \approx f(x)$$
for $x$ near $c$.
The taylor series is formed by taking $T_\infty$.
A: I bet the OP was asking to find the values of the derivatives in $x=1$.
This is clearly the same as finding the derivatives in $x=1$ for $f(x)=\frac{1}{x^2+1}$. Since:
$$\frac{1}{x^2+1}=\frac{1}{2i}\left(\frac{1}{x-i}-\frac{1}{x+i}\right),\tag{1}$$
we have that:
$$\frac{d^n}{dx^n}\frac{1}{x^2+1}=\frac{n!(-1)^n}{2i}\left(\frac{1}{(x-i)^{n+1}}-\frac{1}{(x+i)^{n+1}}\right)\tag{2}$$
so:
$$ f^{(n)}(1) = n!(-1)^n\cdot \Im\left(\frac{1}{(1-i)^{n+1}}\right)\tag{3}$$
and since $(1-i)=\sqrt{2}e^{-i\pi/4}$ we have:
$$ f^{(n)}(1) = n! (-1)^n \sin\left(\frac{\pi(n+1)}{4}\right)2^{-\frac{n+1}{2}}\tag{4}$$
and:
$$ \frac{1}{x^2+1}=\sum_{k=0}^{+\infty}\frac{f^{(k)}(1)}{k!}(x-1)^k = \sum_{k=0}^{+\infty} (-1)^k \sin\left(\frac{\pi(k+1)}{4}\right)2^{-\frac{k+1}{2}}(x-1)^k.\tag{5} $$
The radius of convergence is given, as usual, by the distance from the closest singularity, the simple pole in $x=i$, giving $\rho=\sqrt{2}$. Integrating termwise we get:
$$ \arctan x = \frac{\pi}{4}+\sum_{k=0}^{+\infty} \frac{(-1)^k}{k+1} \sin\left(\frac{\pi(k+1)}{4}\right)2^{-\frac{k+1}{2}}(x-1)^{k+1}.\tag{6} $$
