<s>[INST] prove $\ln(1+x^2)\arctan x=-2\sum_{n=1}^\infty \frac{(-1)^n H_{2n}}{2n+1}x^{2n+1}$ I was able to prove the above identity using 1) Cauchy Product of Power series and 2) integration but the point of posting it here is to use it as a reference in our solutions. 
other approaches would be appreciated. 
 [/INST] I think this is a much simpler proof.
\begin{align}
\tanh^{-1}x\ln(1-x^2)&=\frac12\{\ln(1+x)-\ln(1-x)\}\{\ln(1+x)+\ln(1-x)\}\tag1\\
&=\frac12\ln^2(1+x)-\frac12\ln^2(1-x)\tag2\\
&=\sum_{n=1}^\infty(-1)^n\frac{H_{n-1}}{n}x^n-\sum_{n=1}^\infty\frac{H_{n-1}}{n}x^n\tag3\\
&=-2\sum_{n=1}^\infty\frac{H_{2n-2}}{2n-1}x^{2n-1}\tag4\\
&=-2\sum_{n=1}^\infty\frac{H_{2n}}{2n+1}x^{2n+1}\tag5
\end{align}
Thus $$\tanh^{-1}x\ln(1-x^2)=-2\sum_{n=1}^\infty\frac{H_{2n}}{2n+1}x^{2n+1}\tag6$$
Replace $x$ with $ix$ we get 
$$\tan^{-1}x\ln(1+x^2)=-2\sum_{n=1}^\infty\frac{(-1)^nH_{2n}}{2n+1}x^{2n+1}\tag7$$

Explanation:
$(1)$ $\tanh^{-1}x=\frac12\ln\left(\frac{1+x}{1-x}\right)$.
$(2)$ Difference of two squares.
$(3)$ $\frac12\ln^2(1-x)=\sum_{n=1}^\infty\frac{H_n}{n+1}x^{n+1}=\sum_{n=1}^\infty\frac{H_{n-1}}{n}x^n$
$(4)$ $\sum_{n=1}^\infty ((-1)^n-1)a_{n}=-2\sum_{n=1}^\infty a_{2n-1}$
$(5)$ Reindex.

Bonus:
If we differentiate both sides of $(7)$ we obtain another useful identity
$$\frac{\arctan x}{1+x^2}=\frac12\sum_{n=1}^\infty(-1)^n\left(H_n-2H_{2n}\right)x^{2n-1}\tag8$$
another identity follows from integrating both sides of $(8)$:
$$\arctan^2x=\frac12\sum_{n=1}^\infty\frac{(-1)^n\left(H_n-2H_{2n}\right)}{n}x^{2n}\tag9$$
replace $x$ with $ix$ in $(9)$
$$\text{arctanh}^2x=-\frac12\sum_{n=1}^\infty\frac{\left(H_n-2H_{2n}\right)}{n}x^{2n}\tag{10}$$
</s>