Let $\mathcal{H}_n$ denote the $n$-th harmonic number. Prove that forall $|x|<1$ it holds that

\[\sum_{n=1}^{\infty} \frac{(-1)^{n-1} \mathcal{H}_{2n} x^{2n+1}}{2n+1} = \frac{\arctan x \log (1+x^2)}{2}\]

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Let the proposed series be $f(x)$. Then
$$f'(x) = \sum_{n=1}^\infty (-1)^{n-1}H_{2n}x^{2n} = - \sum_{n=1}^\infty H_{2n}(-x^2)^n.$$
Recall that
$$\sum_{n=1}^\infty H_nz^n = - \frac{\ln(1-z)}{1-z}.$$
It follows that
\begin{eqnarray*}
\sum_{n=1}^\infty H_{2n}z^{2n} & = & \frac{1}{2}\,(\sum_{n=1}^\infty H_nz^n + \sum_{n=1}^\infty H_n(-z)^n)\\
& = & - \frac{1}{2}\left(\frac{\ln(1-z)}{1-z} + \frac{\ln(1+z)}{1+z}\right).
\end{eqnarray*}
Setting $z = ix$ gives
\begin{eqnarray*}
\sum_{n=1}^\infty H_{2n}(-x^2)^n & = & - \frac{1}{2}\left(\frac{\ln(1-ix)}{1-ix} + \frac{\ln(1+ix)}{1+ix}\right)\\
& = & - \frac{1}{2}\,\frac{2x\arctan x+ \ln(1+x^2)}{1 + x^2},
\end{eqnarray*}
where we have used $\ln(1\pm ix) = \frac{1}{2}\ln(1 + x^2) \pm i\arctan x$. Thus,
$$f'(x) = \frac{1}{2}\,\frac{2x\arctan x+ \ln(1+x^2)}{1 + x^2}.$$
Integrating with respect to $x$ yields
$$f(x) = \frac{1}{2}\,\arctan x\ln(1+x^2)$$
as desired.