Express $\arctan(2x)$ as power series

Question

I am trying to express $\arctan(2x)$ as power series.

Let $f(x) = \arctan(2x)$, then $f'(x) = \frac{2}{1+4x^2}$

$$\arctan(2x)=\int \frac{2}{1+4x^2} \, dx$$

$$\int\frac{2}{1+4x^2} \, dx= \int2\frac{1}{1-(-4x^2)} \, dx =\int \sum_{n=0}^\infty 2(-1)^n (4x^2)^n \, dx$$ (geometric series)

I feel like I am doing something wrong here and would like to get some feedback

Answer 1

Following your steps, one has $$ \frac{2}{1+4x^{2}}= 2\frac{1}{1-(-4x^{2})} = \sum_{n=0}^{\infty}(-1)^{n}2^{2n+1}x^{2n}, \quad |x|<\frac12, $$ we are allowed to integrate termwise as long as $|x|<\dfrac12$ obtaining $$ \arctan(2x)=\int_0^x\frac{2}{1+4t^{2}}\:dt=\sum_{n=0}^{\infty}(-1)^{n}2^{2n+1}\frac{x^{2n+1}}{2n+1}, \quad |x|<\frac12. $$

Answer 2

The geometric series that you got is for $\dfrac d {dx} \arctan(2x)$, not for $\arctan(2x)$. You need to take an antiderivative of that.