Calculate the series expansion at $x=0$ of the integral $\int \frac{xy\arctan(xy)}{1-xy}dx$ When I was studying $$\int_0^1\int_0^1\frac{x^\alpha y^\beta\arctan(xy)}{1-xy}dxdy,$$  (I tried get the case $\alpha=\beta$ and the general case with a Cauchy product) I've asked to Wolfram Alpha online calculator a simple case
integrate xy arctan(xy)/(1-xy)dx
and one of the outputs was the series expansion of the integral at $x=0$.

Question. Can you explain me how do you get the series expansion of the integral 
  $$\int \frac{x\arctan(xy)}{1-xy}dx$$ at $x=0$? Thanks in advance.

 A: Consider 
$$\arctan(x) = \int^x_0 \frac{1}{1+t^2}\,dx = \sum_{k=0}^\infty (-1)^k \frac{x^{2k+1}}{2k+1}$$
We can rewrite as 
$$\sum_{k=0}^\infty a_k x^k$$
Where $a_{2j}=0$ and  $a_{2j+1} = \frac{(-1)^j}{2j+1}$ , also we have 
$$\frac{1}{1-x} = \sum_{k=0}^\infty b _kx^k$$
where $b_k = 1$
Then by Cauchy product formula we have 
$$\left(\sum_{k=0}^\infty a_k x^k\right)\left(\sum_{k=0}^\infty b_k x^k\right)
=\sum_{k=0}^\infty\left(\sum_{j=0}^k a_jb_{k-j}\right)x^k$$
We need to find 
$$\sum_{j=0}^k a_jb_{k-j} = \sum_{j=0}^k a_j$$
Consider the two cases when $k =2n$ is even 
$$ \sum_{j=0}^{2n} a_j = \sum_{j=0}^{n} a_{2j}+ \sum_{j=0}^{n-1} a_{2j+1} = \sum_{j=0}^{(k/2)-1} \frac{(-1)^j}{2j+1}$$
If $k = 2n+1$ is odd then 
$$ \sum_{j=0}^{2n+1} a_j = \sum_{j=0}^{n-1} a_{2j}+ \sum_{j=0}^{n} a_{2j+1} = \sum_{j=0}^{(k-1)/2} \frac{(-1)^j}{2j+1}$$
Hence we know that 
$$\sum_{j=0}^k a_j = \sum_{j=0}^{\lceil \frac{k}{2} \rceil -1} \frac{(-1)^j}{2j+1} $$
Finally we have 

$$\frac{\arctan(xy)}{1-xy} = \sum_{k=0}^\infty \left(\sum_{j=0}^{\lceil
 \frac{k}{2} \rceil -1} \frac{(-1)^j}{2j+1} \right) (xy)^k $$

Note that the coefficients are variations of the alternating Harmonic numbers. 
