I will show that
$$\int_0^1 \frac{1 - x}{1 + x} \frac{x^n}{\sqrt{x^4 - 2x^2 + 1}} \, dx = -\frac{1}{2} + n (-1)^{n + 1} \left (\ln 2 + \sum_{k = 1}^{n - 1} \frac{(-1)^k}{k} \right ), n \geqslant 1.$$
Note here we interpret the empty sum as being equal to zero (the empty sum is the case when $n = 1$ in the finite sum) and I assume $n \in \mathbb{N}$.
For the case when $n = 0$, a direct evaluation yields: $I_0 = \frac{1}{2}$.
As already noted, since
$$\frac{1-x}{1+x}\cdot \frac{x^n}{\sqrt{x^4-2x^2+1}}=\dfrac{x^n}{(1+x)^2},$$
the integral becomes
$$I_n = \int_0^1 \frac{x^n}{(1 + x)^2} \, dx.$$
For $n \in \mathbb{N}$, integrating by parts we have
\begin{align}
I_n &= \left [-\frac{x^n}{1 + x} \right ]_0^1 + n\int_0^1 \frac{x^{n - 1}}{1 + x} \, dx\\
&= -\frac{1}{2} + n \int_0^1 \frac{x^{n - 1}}{1 + x} \, dx\\
&= -\frac{1}{2} + n \sum_{k = 0}^\infty (-1)^k \int_0^1 x^{n + k - 1} \, dx\\
&= -\frac{1}{2} + n \sum_{k = 0}^\infty \frac{(-1)^k}{n + k}.
\end{align}
Note here the geometric sum for $1/(1 + x)$ of $\sum_{k = 0}^\infty (-1)^k x^k$ has been used. Reindexing the sum $k \mapsto k - n$ gives
\begin{align}
I_n &= -\frac{1}{2} + n (-1)^n \sum_{k = n}^\infty \frac{(-1)^k}{k}\\
&= -\frac{1}{2} + n(-1)^n \left (\sum_{k = 1}^\infty \frac{(-1)^k}{k} - \sum_{k = 1}^{n - 1} \frac{(-1)^k}{k} \right )\\
&= -\frac{1}{2} + n(-1)^{n + 1} \left (\ln 2 + \sum_{k = 1}^{n - 1} \frac{(-1)^k}{k} \right ),
\end{align}
where I have made use of the well-known result of $\ln 2 = -\sum_{k = 1}^\infty (-1)^k/k$.
In terms of your $H^*_n$ notation for the finite sum $\sum_{k = 1}^n \frac{(-1)^{k + 1}}{k}$, this result can be re-expressed as
$$I_n = -\frac{1}{2} + n (-1)^{n + 1} (\ln 2 - H^*_{n - 1}).$$
To show my result is equivalent to the result you quote, one would need to show, after playing around with finite sums, that
$$\sum_{k = 1}^{n - 1} H^*_k = -\frac{1}{2} (1 + (-1)^n) + n H^*_{n - 1}.$$
