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

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 $I_0=?,I_1=?$
