Prove $\int_0^1\frac{x^{2n}}{1+x}dx=\ln2+H_n-H_{2n}$

\begin{align} \int_0^1\frac{x^{2n}}{1+x}\mathrm{d}x &=\int_0^1x^{2n}\sum_{k=0}^\infty(-x)^k\mathrm{d}x\\ &=\sum_{k=0}^\infty(-1)^k\int_0^1x^{2n+k}\mathrm{d}x\\ &=\sum_{k=0}^\infty\frac{(-1)^k}{2n+k+1}\\ &=\sum_{j=2n+1}^\infty\frac{(-1)^{j+1}}{j}\\ &=\sum_{j=1}^\infty\frac{(-1)^{j+1}}j-\sum_{j=1}^{2n}\frac{(-1)^{j+1}}j\\ &=\ln{(2)}+H_n-H_{2n}\\ \end{align}

We have using only integration of rational functions: $$ \begin{aligned} \int_0^1 \frac{x^{2n}}{x+1}\; dx &= \int_0^1 \frac{x^{2n}+x}{x+1}\; dx - \int_0^1 \frac{x}{x+1}\; dx \\ &= \int_0^1 \Big(x^{2n-1}-x^{2n-2}+\dots- x^4 + x^3 - x^2 + x\Big)\; dx - \int_0^1 \frac{x}{x+1}\; dx \\ &= \left(\frac 1{2n}-\frac 1{2n-1}+\dots -\frac 15+\frac 14-\frac 13+\frac 12\right)-1+\log 2 \\ &= \log 2 - H_{2n}+2\left( \frac 12+\frac 14+\dots+\frac 1{2n}\right) \\ &= \log 2 - H_{2n}+H_n\ . \end{aligned} $$