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

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
$$\int_0^1\frac{dx}{(x+1)^2}=\left.-\frac1{(x+1)}\right|_0^1=-\frac12+1=\frac12$$
