How do I find a Taylor series of $\arctan \frac{2-2x}{1+4x}$ at $x=0$

Question

I've been trying to find a Taylor series of $\arctan \frac{2-2x}{1+4x}$ at $x=0$ The only thing I could think of was trying to find formula for the $n$th derivative but was unable to find it so it would fit into the result.

Answer 1

Hint:

$$\arctan\dfrac{2-2x}{1+4x}=\arctan\dfrac{2+(-2x)}{1-2(-2x)}=?$$

See Inverse trigonometric function identity doubt: $\tan^{-1}x+\tan^{-1}y =-\pi+\tan^{-1}\left(\frac{x+y}{1-xy}\right)$, when $x<0$, $y<0$, and $xy>1$

Answer 2

Another hint:

The derivative is $$\biggl(\arctan\frac{2-2x}{1+4x}\biggr)'=-\frac 2{1+4x^2}=-2\sum_{k=0}^\infty(-1)^n 4^n x^{2n},$$ so integrating, we get $$\arctan\frac{2-2x}{1+4x}-\arctan 2= \sum_{k=0}^\infty(-1)^{n+1} 4^{n+1}\frac{x^{2n+1}}{2n+1}.$$