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. 

actually, you can do the product directly, given well-known series
$$\begin{aligned}
\arctan x & = \sum_{n=0}^{\infty} {\frac{(-1)^n x^{2n+1}}{2n+1}}\\
\ln(1+x^2) & = \sum_{n=1}^{\infty} {\frac{(-1)^{n+1} x^{2n}}{n}}
\end{aligned}$$
obviously their product has no even order items, set
$$\arctan x \ln (1+x^2) = \sum_{m=0}^{\infty} {a_{2m+1} x^{2m+1}}$$
for item $x^{2m+1}$, it has pair partitions as $(x,x^{2m}),(x^3,x^{2m-2}),\cdots,(x^{2m-1},x^2)$, thus
$$\begin{aligned}
a_{2m+1} 
& = \sum_{n=0}^{m-1} {\frac{(-1)^n}{2n+1} \cdot \frac{(-1)^{m-n+1}}{m-n}} = \sum_{n=0}^{m-1} {\frac{(-1)^{m+1}}{(2n+1)(m-n)}}\\
& = \frac{(-1)^{m+1}}{2m+1} \sum_{n=0}^{m-1} {\frac{2m+1}{(2n+1)(m-n)}} = \frac{(-1)^{m+1}}{2m+1} \sum_{n=0}^{m-1} {\frac{2n+1+2(m-n)}{(2n+1)(m-n)}}\\
& = \frac{(-1)^{m+1}}{2m+1} \left( \sum_{n=0}^{m-1} {\frac1{m-n}} + \sum_{n=0}^{m-1} {\frac2{2n+1}} \right)\\
& = \frac{(-1)^{m+1}}{2m+1} \left( H_{m} + 2\left( \sum_{n=1}^{2m} {\frac1{n}} - \sum_{n=1}^{m} {\frac1{2n}} \right) \right)\\
& = \frac{(-1)^{m+1} (H_{m} + 2H_{2m} - H_{m})}{2m+1} = \frac{(-1)^{m+1} \cdot 2H_{2m}}{2m+1}
\end{aligned}$$
