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.

We can try to obtain the series in the following way:
$$f(x)=\arctan^2 x=x^2 \int_0^1 \int_0^1 \frac{du~dv}{(1+x^2u^2)(1+x^2v^2)}$$
It's easier to consider:
$$g(x)=\int_0^1 \int_0^1 \frac{du~dv}{(1+x^2u^2)(1+x^2v^2)}$$
Let's use partial fractions:
$$\frac{1}{(1+x^2u^2)(1+x^2v^2)}=\frac{u^2}{(u^2-v^2)(1+x^2u^2)}-\frac{v^2}{(u^2-v^2)(1+x^2v^2)}$$
We obtain a sum of two singular integrals, which however, can both be formally expanded into a series:
$$g(x)=\int_0^1 \int_0^1 \left(\frac{u^2}{(u^2-v^2)(1+x^2u^2)}-\frac{v^2}{(u^2-v^2)(1+x^2v^2)} \right) du ~dv=$$
$$g(x)=\sum_{n=0}^\infty (-1)^n x^{2n} \int_0^1 \int_0^1 \left(\frac{u^{2n+2}}{u^2-v^2}-\frac{v^{2n+2}}{u^2-v^2} \right) du ~dv$$
Now:
$$g(x)=\sum_{n=0}^\infty (-1)^n x^{2n} \int_0^1 \int_0^1 \frac{u^{2n+2}-v^{2n+2}}{u^2-v^2} du ~dv$$
Obviously, every integral is finite now, and we can write:
$$\int_0^1 \int_0^1 \frac{u^{2n+2}-v^{2n+2}}{u^2-v^2} dv ~du=2 \int_0^1 \int_0^u \frac{u^{2n+2}-v^{2n+2}}{u^2-v^2} dv ~du= \\ = 2 \sum_{k=0}^\infty \int_0^1 \int_0^u u^{2n} \left(1-\frac{v^{2n+2}}{u^{2n+2}} \right) \frac{v^{2k}}{u^{2k}} dv ~du =2 \sum_{k=0}^\infty \int_0^1 \int_0^1 u^{2n+1} \left(1-t^{2n+2} \right) t^{2k} dt ~du= \\ = 2 \sum_{k=0}^\infty \int_0^1 \left(\frac{1}{2k+1}-\frac{1}{2k+2n+3} \right) u^{2n+1}~du= 2\sum_{k=0}^\infty \frac{1}{(2k+1)(2k+2n+3)}$$
So we get:
$$g(x)=\frac{1}{2} \sum_{n=0}^\infty (-1)^n x^{2n} \sum_{k=0}^\infty \frac{1}{(k+\frac{1}{2})(k+n+\frac{3}{2})}$$
The inner series converges for all $n$, and we can formally represent it as a difference of two divergent harmonic series, which, after some manipulations, should give us the same result as robjohn obtained.
I suppose, a kind of closed form for the general term can also be given in terms of digamma function:
$$g(x)=\frac{1}{2} \sum_{n=0}^\infty (-1)^n \frac{\psi \left(n+\frac32 \right)-\psi \left(\frac12 \right)}{n+1} x^{2n}$$
Which makes:
$$\arctan^2 x=\frac{x^2}{2} \sum_{n=0}^\infty (-1)^n \frac{\psi \left(n+\frac32 \right)-\psi \left(\frac12 \right)}{n+1} x^{2n}$$
Which is essentially the same as robjohn's answer.
