How to evaluate $\int_{0}^{1}\frac{1-x}{1+x}\cdot \frac{x^n}{\sqrt{x^4-2x^2+1}}\mathrm dx$ I am trying to calculate this integral, but I find it is very challenging
$$\int_{0}^{1}\frac{1-x}{1+x}\cdot \frac{x^n}{\sqrt{x^4-2x^2+1}}\mathrm dx$$
but somehow I have managed to local its closed form to be $$(-1)^n\left(\frac{1}{2}-n\ln 2+\sum_{j=0}^{n-1}H^{*}_j\right)$$ 
where $n\ge 0$, $H^{*}_0=0$ and $H^{*}_k=\sum_{j=1}^{k}\frac{(-1)^{j+1}}{j}$
I have try
$$\frac{1-x}{1+x}\cdot \frac{x^n}{(x^2-1)^{1/2}}$$
$$-x^n\sqrt{\frac{x-1}{(x+1)^2}}$$
$$-\int_{0}^{1}x^n\sqrt{\frac{x-1}{(x+1)^3}}\mathrm dx$$
from this point I tried to use the binomial to expand $$\sqrt{\frac{x-1}{(x+1)^3}}$$ but it seem not possible
 A: In $0\le x<1$
$$\frac{1-x}{1+x}\cdot \frac{x^n}{\sqrt{x^4-2x^2+1}}=\dfrac{x^n}{(1+x)^2}$$
If $\displaystyle I_n=\int_{0}^{1}\dfrac{1-x}{1+x}\cdot \frac{x^n}{\sqrt{x^4-2x^2+1}}\mathrm dx$
$$I_n+2I_{n-1}+I_{n-2}=\int_0^1x^{n-2} dx$$
Now $I_0=?,I_1=?$
A: 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
$$\sum_{k = 1}^{n - 1} H^*_k = -\frac{1}{2} (1 + (-1)^n) + n H^*_{n - 1}.$$
A: My solution is similar to that of @omegadot but I think it may be a little cleaner. For $n\in\mathbb{Z}^+$ let
$$C_n=(-1)^n\int_0^1\frac{x^{n-1}}{x+1}dx$$
Then
$$C_1=-\int_0^1\frac{dx}{x+1}=\left.-\ln(x+1)\right|_0^1=-\ln2$$
$$C_{n+1}-C_n=(-1)^{n+1}\int_0^1\frac{x^n+x^{n-1}}{x+1}dx=(-1)^{n+1}\int_0^1x^{n-1}dx=\frac{(-1)^{n+1}}n$$
So we can sum a telescoping series to get
$$C_n-C_1=\sum_{k=1}^{n-1}\left(C_{k+1}-C_k\right)=\sum_{k=1}^{n-1}\frac{(-1)^{k+1}}k$$
Then since
$$\sqrt{x^4-2x^2+1}=\sqrt{\left(1-x^2\right)^2}=\left|1-x^2\right|=1-x^2=(1+x)(1-x)$$
for $0\le x\le 1$, we have for $n\in\mathbb{Z}^+$
$$\begin{align}\int_0^1\frac{1-x}{1+x}\cdot\frac{x^n}{\sqrt{x^4-2x^2+1}}dx&=\int_0^1\frac{x^n}{(x+1)^2}dx=\left.-\frac1{(x+1)}x^n\right|_0^1+n\int_0^1\frac{x^{n-1}}{x+1}dx\\
&=-\frac12+(-1)^nnC_n\\
&=-\frac12+(-1)^nn\left[-\ln 2-\sum_{k=1}^{n-1}\frac{(-1)^k}k\right]\end{align}$$
This is using a recurrence relation rather than an infinite series. Of course for $n=0$ the integral works out to
$$\int_0^1\frac{dx}{(x+1)^2}=\left.-\frac1{(x+1)}\right|_0^1=-\frac12+1=\frac12$$
