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. 

knowing that fact that
$$2\sum_{n=1}^\infty f(2n)=\sum_{n=1}^\infty f(n)(1+(-1)^n)$$
then 
\begin{align}
2\sum_{n=1}^\infty (-1)^nx^{2n}H_{2n}&=2\sum_{n=1}^\infty (i)^{2n}x^{2n}H_{2n}\\
&=\sum_{n=1}^\infty (ix)^nH_{n}+\sum_{n=1}^\infty (-ix)^nH_{n}\\
&=-\frac{\ln(1-ix)}{1-ix}-\frac{\ln(1+ix)}{1+ix}\\
&=-\frac{\ln(1-ix)+\ln(1+ix)+ix(\ln(1-ix)-\ln(1+ix))}{1+x^2}\\
&=-\frac{\ln(1+x^2)+ix(-2i\arctan x)}{1+x^2}\\
&=-\frac{\ln(1+x^2)}{1+x^2}-\frac{2x\arctan x}{1+x^2}
\end{align}
integrate both sides from $x=0$ to $z$
\begin{align}
2\sum_{n=1}^\infty (-1)^nH_{2n}\int_0^zx^{2n}\ dx&=2\sum_{n=1}^\infty\frac{(-1)^nH_{2n}}{2n+1}z^{2n+1}\\
&=-\int_0^z\left(\frac{\ln(1+x^2)}{1+x^2}+\frac{2x\arctan x}{1+x^2}\right)\ dx\\
&=-\int_0^zd(\ln(1+x^2)\arctan x)\\
&=-\ln(1+z^2)\arctan z
\end{align}
