Limit of $\int_0^1 \frac{x^n-x^{2n}}{1-x}\text{d}x$

Just with Riemann sums:

First rewrite the integrand as $$ \frac{x^n-x^{2n}}{1-x}=x^n\frac{1-x^{n}}{1-x}=x^n(1+x+\dots+x^{n-1}) =\sum_{k=0}^{n-1}x^{n+k}, $$ whence the value of the integral $$\int_0^1\frac{x^n-x^{2n}}{1-x}\,\mathrm dx=\sum_{k=1}^{n}\frac1{n+k} =\frac1n\sum_{k=1}^{n}\frac1{1+\cfrac kn}.$$ Now the latter expression is the lower Riemann sum for the integral $$\int_0^1\frac{\mathrm dx}{1+x}=\ln(1+x)\Big\vert_0^1.$$

Your integral is $H_{2n}-H_n$ which using known behavior of $H_n$ at large $n$, that is $H_{n\gg 1} = \ln n + C$, gives $$\lim_{n\to\infty}I=\lim_{n\to\infty}(H_{2n}-H_{n})=\ln 2.$$