How to prove
$$\int_0^1\frac{x^{2n}}{1+x}dx=\ln2+H_n-H_{2n}$$
I used this identity to solve some advanced harmonic series but I didn't provide a proof so I see that it's worth a post so that we can use it as a reference for future solutions if needed. Here is my approach and would like to see alternative ones.
\begin{align} \int_0^1\frac{x^{2n}}{1+x}dx&=\ln2-2n\int_0^1x^{2n-1}\ln(1+x)dx\tag1\\ &=\ln2-2n\sum_{k=1}^\infty\frac{(-1)^{k-1}}{k}\int_0^1 x^{2n+k-1}dx\tag2\\ &=\ln2+2n\sum_{k=1}^\infty\frac{(-1)^{k}}{k(k+2n)}\tag3\\ &=\ln2+4n\sum_{k=1}^\infty\frac{1}{2k(2k+2n)}-2n\sum_{k=1}^\infty\frac{1}{k(k+2n)}\tag4\\ &=\ln2+\sum_{k=1}^\infty\frac{n}{k(k+n)}-\sum_{k=1}^\infty\frac{2n}{k(k+2n)}\tag5\\ &=\ln2+H_n-H_{2n}\tag6 \end{align}
Explanation:
1) Apply integration by parts
2) Write $\ln(1+x)=\sum_{k=1}^\infty \frac{(-1)^{k-1}}{k}x^{k}$
3) Use the rule $\int_0^1 x^ndx=\frac1{n+1}$
4) $\sum_{k=1}^\infty (-1)^k f(k)=2\sum_{k=1}^\infty f(2k)-\sum_{k=1}^\infty f(k)$
5) Simplify
6) Use $H_n=\sum_{k=1}^n \frac1k=\sum_{k=1}^\infty\frac{n}{k(k+n)}$
A good application for this identity is the following problem proposed by Cornel:
$$\zeta(3)=\frac43\sum_{n=1}^\infty\frac{(2H_{2n}-H_n)(H_n-H_{2n}+\ln2)}{n}$$
If we multiply both sides of our identity by $\frac{2H_{2n}-H_n}{n}$ then sum up from $n= 1$ to $\infty$ we get
$$\sum_{n=1}^\infty\frac{(2H_{2n}-H_n)(H_n-H_{2n}+\ln2)}{n}=\int_0^1\frac1{1+x}\sum_{n=1}^\infty\frac{x^{2n}}{n}(2H_{2n}-H_n)dx\\=\frac12\int_0^1\frac{1}{1+x}\ln^2\left(\frac{1-x}{1+x}\right)dx=\frac12\int_0^1\frac{\ln^2x}{1+x}dx=\frac34\zeta(3)$$
where the identity $\ln^2\left(\frac{1-x}{1+x}\right)=2\sum_{n=1}^\infty \frac{x^{2n}}{n}(2H_{2n}-H_n)$ was used in our calculations.
Another application is calculating $\sum_{n=1}^\infty \frac{(-1)^nH_{n/2}}{n^3}$:
From our proof above, we can see that
$$\int_0^1 x^{2n-1}\ln(1+x)dx=\frac{H_{2n}-H_n}{2n}$$
Replace $2n$ by $n$ then multiply both sides by $\frac{(-1)^n}{n^2}$ and sum up we get
$$\sum_{n=1}^\infty \frac{(-1)^nH_n}{n^3}-\sum_{n=1}^\infty \frac{(-1)^nH_{n/2}}{n^3}=\int_0^1\frac{\ln(1+x)}{x}\sum_{n=1}^\infty \frac{(-x)^n}{n^2}dx\\=\int_0^1\frac{\ln(1+x)\operatorname{Li}_2(-x)}{x}dx=-\frac12\operatorname{Li}_2^2(-1)=-\frac12\left(-\frac12\zeta(2)\right)^2=-\frac5{16}\zeta(4)$$
I managed here to prove
$$\sum_{n=1}^\infty \frac{(-1)^nH_n}{n^3}=2\operatorname{Li_4}\left(\frac12\right)-\frac{11}4\zeta(4)+\frac74\ln2\zeta(3)-\frac12\ln^22\zeta(2)+\frac{1}{12}\ln^42$$
Thus
$$\sum_{n=1}^\infty \frac{(-1)^nH_{n/2}}{n^3}=2\operatorname{Li_4}\left(\frac12\right)-\frac{39}{16}\zeta(4)+\frac74\ln2\zeta(3)-\frac12\ln^22\zeta(2)+\frac{1}{12}\ln^42$$
