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
Best Answer
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.