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. 
 A: 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
$$\begin{aligned}
a_{2m+1} 
& = \sum_{n=0}^{m-1} {\frac{(-1)^n}{2n+1} \cdot \frac{(-1)^{m-n+1}}{m-n}} = \sum_{n=0}^{m-1} {\frac{(-1)^{m+1}}{(2n+1)(m-n)}}\\
& = \frac{(-1)^{m+1}}{2m+1} \sum_{n=0}^{m-1} {\frac{2m+1}{(2n+1)(m-n)}} = \frac{(-1)^{m+1}}{2m+1} \sum_{n=0}^{m-1} {\frac{2n+1+2(m-n)}{(2n+1)(m-n)}}\\
& = \frac{(-1)^{m+1}}{2m+1} \left( \sum_{n=0}^{m-1} {\frac1{m-n}} + \sum_{n=0}^{m-1} {\frac2{2n+1}} \right)\\
& = \frac{(-1)^{m+1}}{2m+1} \left( H_{m} + 2\left( \sum_{n=1}^{2m} {\frac1{n}} - \sum_{n=1}^{m} {\frac1{2n}} \right) \right)\\
& = \frac{(-1)^{m+1} (H_{m} + 2H_{2m} - H_{m})}{2m+1} = \frac{(-1)^{m+1} \cdot 2H_{2m}}{2m+1}
\end{aligned}$$
A: 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}
A: 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 $(9)$
$$\text{arctanh}^2x=-\frac12\sum_{n=1}^\infty\frac{\left(H_n-2H_{2n}\right)}{n}x^{2n}\tag{10}$$
