Alternative ways to evaluate $\int_0^1 x^{2n-1}\ln(1+x)dx =\frac{H_{2n}-H_n}{2n}$ For all, $n\geq 1$, prove that

$$\int_0^1 x^{2n-1}\ln(1+x)\,dx =\frac{H_{2n}-H_n}{2n}\tag1$$
This identity I came across to know from here, YouTube which is proved in elementary way.

Trying to make alternate efforts to prove  it, we can observe  that $$\ln(1+x)=\ln\left(\frac{1-x^2}{1-x}\right)=\ln(1-x^2)-\ln(1-x)$$On multiplying by $x^{2n-1}$ and integrating from $0$ to $1$ , we have first order partial derivatives of beta function.
How can we prove that identity $(1)$ without the use of derivatives of beta function?
 A: An alternative  approach
It is well know that around $x=0$   function, $\ln(1+x)$ admits expansion  $\displaystyle=\sum_{k\geq 1}(-1)^{k+1}\frac{x^k}{k}$. So exploiting the result and hence on interchange of summation and integral signs  reduces to $$\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k}\int_0^1 x^{2n+k-1}dx =\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k(k+2n)}=\sum_{k=1}^{\infty}\left(\frac{1}{(2k-1)(2n+2k-1)}-\frac{1}{2k(2n+2k)}\right)=\frac{1}{2n}\sum_{k=1}^{\infty}\left(\frac{1}{2k-1}-\frac{1}{2k+2n-1}\right)-\frac{1}{4n}\sum_{k=1}^{\infty}\frac{n}{k(n+k)}=\frac{1}{4n}\left(\psi_0\left(n+\frac{1}{2}\right)-\psi_0\left(\frac{1}{2}\right)-\psi_0(n+1)+\gamma\right)$$we obtain the latter result by series definition of digamma function and by the use of   half digamma integer argument we can have $\displaystyle \psi_0\left(\frac{1}{2}\right)=-\gamma -2\ln 2$ and thus latter expression simplies  to $$\frac{1}{4n}\left(\psi_0\left(n+\frac{1}{2}\right)+\gamma+2\ln2 -\psi_0(n+1)+\psi\right)=\frac{1}{4n}\left(H_{n-\frac{1}{2}}+2\ln 2 -H_n\right)=\frac{H_{2n}-H_n}{2n}$$ since $\displaystyle H_{n-\frac{1}{2}} =2H_{2n} -H_n-2\ln 2 $due to multiplication formula of harmonic number.
Other ways are most welcome.
A: It is not necessary to use digamma function and multiplication formula of harmonic numbers, indeed
$$\int_0^1 x^{2n-1}\ln(1+x)dx=\\=\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k}\int_0^1 x^{2n+k-1}dx =\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k(k+2n)}=\\=\sum_{k=1}^{\infty}\left(\frac{1}{(2k-1)(2n+2k-1)}-\frac{1}{2k(2n+2k)}\right)=\\=\frac{1}{2n}\sum_{k=1}^{\infty}\left(\frac{1}{2k-1}-\frac{1}{2(n+k)-1}\right)-\frac{1}{4n}\sum_{k=1}^{\infty}\frac{n}{k(n+k)}=\\=\frac{1}{2n}\sum_{k=1}^n\frac{1}{2k-1}-\frac{1}{4n}\sum_{k=1}^{\infty}\left(\frac{1}{k}-\frac{1}{n+k}\right)=\\=\frac{1}{2n}\sum_{k=1}^n\frac{1}{2k-1}-\frac{1}{4n}\sum_{k=1}^n\frac{1}{k}=\\=\frac{1}{2n}\left(\sum_{k=1}^n\frac{1}{2k-1}-\sum_{k=1}^n\frac{1}{2k}\right)=\\=\frac{1}{2n}\left(\sum_{k=1}^n\frac{1}{2k-1}+\sum_{k=1}^n\frac{1}{2k}-\sum_{k=1}^n\frac{1}{k}\right)=\\=\frac{1}{2n}\left(\sum_{h=1}^{2n}\frac{1}{h}-\sum_{k=1}^n\frac{1}{k}\right)=\frac{H_{2n}-H_n}{2n}\;.$$
A: Simply integrate by parts: $$2n\int_0^1 x^{2n-1}\ln(1+x)\,dx=x^{2n}\ln(1+x)\Bigg|_0^1-\int_0^1\frac{x^{2n}}{1+x}\,dx=\int_0^1\frac{1-x^{2n}}{1+x}\,dx.$$ Now use $\displaystyle\frac{1-x^{2n}}{1+x}=(1-x)\frac{1-x^{2n}}{1-x^2}=(1-x)\sum_{k=1}^n x^{2(k-1)}$. You get, as expected, $$\ldots=\sum_{k=1}^n\left(\frac{1}{2k-1}-\frac{1}{2k}\right)=\sum_{k=1}^{2n}\frac1k-2\sum_{k=1}^n\frac{1}{2k}=H_{2n}-H_n.$$
A: Solution by Cornel:
$$\int_0^1 x^{2n-1}\ln(1+x)dx=\underbrace{\int_0^1 x^{2n-1}\ln(1-x^2)dx}_{x^2\to x}-\int_0^1 x^{2n-1}\ln(1-x)dx$$
$$=\frac12\int_0^1 x^{n-1}\ln(1-x)dx-\int_0^1 x^{2n-1}\ln(1-x)dx$$
$$=\frac12\left(-\frac{H_n}{n}\right)-\left(-\frac{H_{2n}}{2n}\right)=\frac{-H_n+H_{2n}}{2n}$$

Note that
$$\int_0^1 x^{n-1}\ln(1-x)dx\overset{IBP}=\underbrace{\left(\frac{x^n}{n}-\frac1n\right)\ln(1-x)\Bigg|_0^1}_{0}+\frac1n\int_0^1\frac{x^n-1}{1-x}dx=-\frac{H_n}{n}$$
