Maclaurin series for $\arctan^{2}(x)$ I have a question here that requires me to find the Maclaurin series expansion of 
$\arctan^{2}(x)$. Now I know how to find it for $\arctan(x)$, by taking the derivative, expanding it into a series, and integrating back (given x is in the interval of uniform convergence), But applying that here leaves me with 
$$\frac{df}{dx}=2\arctan(x)\frac{1}{(x^2+1)}$$ I am not sure If I can pick out parts of the product (like $\arctan(x)$) and differentiate, expand , then integrate them. Even if I did, how would I go about multiplying two series? Is there an easier way to do this? Thanks.

For the sake of an alternative method, here's a double series.
For $|x|<1$, we have 
$$\arctan x=\sum_{k\geq0}\frac{(-1)^kx^{1+2k}}{1+2k}$$
As you noted, 
$$\arctan^2x=2\int\frac{\arctan x}{x^2+1}\mathrm{d}x$$
So, assuming $|x|<1$, 
$$\arctan^2x=2\int\frac1{x^2+1}\sum_{k\geq0}\frac{(-1)^kx^{1+2k}}{1+2k}\mathrm{d}x$$
$$\arctan^2x=2\sum_{k\geq0}\frac{(-1)^k}{1+2k}\int\frac{x^{2k}}{1+x^2}x\mathrm{d}x$$
Now we focus on 
$$I=\int\frac{x^{2k+1}}{1+x^2}\mathrm{d}x$$
We recall that for $|x|<1$, 
$$\frac1{1+x^2}=\frac12\sum_{n\geq0}i^n\big(1+(-1)^n\big)x^n$$
Hence 
$$I=\frac12\sum_{n\geq0}i^n\big(1+(-1)^n\big)\int x^{2k+n+1}\mathrm{d}x$$
$$I=\frac12\sum_{n\geq0}\frac{i^n\big(1+(-1)^n\big)}{2k+n+2}x^{2k+n+2}$$
Then we have our (pretty inefficient) result:
$$\arctan^2x=\sum_{k\geq0}\frac{(-1)^k}{1+2k}\sum_{n\geq0}\frac{i^n\big(1+(-1)^n\big)}{2k+n+2}x^{2k+n+2}$$
$$\arctan^2x=\sum_{n,k\in\Bbb N}\frac{(-1)^{k+\frac{n}2}\big(1+(-1)^n\big)}{(1+2k)(2+2k+n)}x^{2k+n+2}$$
