where $\ \displaystyle\text{B}(n,m)=\int_0^1 x^{n-1}(1-x)^{m-1}\ dx=\frac{\Gamma(n)\Gamma(m)}{\Gamma(n+m)}\ $is the beta function, defined over positive $\ n,m>0$.
The point of this post is to provide a proof for $\ \displaystyle\frac{\partial}{\partial m}\text{B}(n,m)=-\text{B}(n,m)\sum_{k=0}^{n-1}\frac{1}{k+m}$ for, $n$ is a positive integer, so that we can use its' applications as a reference in our solutions and here is some of the applications:
$$\int_0^1x^{n-1}\ln(1-x)\ dx=-\frac{H_n}{n}$$ $$\int_0^1x^{n-1}\ln^2(1-x)\ dx=\frac{H_n^2+H_n^{(2)}}{n}$$ $$\int_0^1x^{n-1}\ln^3(1-x)\ dx=-\frac{H_n^3+3H_nH_n^{(2)}+2H_n^{(3)}}{n}$$ $$\int_0^1x^{n-1}\ln^4(1-x)\ dx=\frac{H_n^4+6H_n^2H_n^{(2)}+8H_nH_n^{(3)}+3\left(H_n^{(2)}\right)^2+6H_n^{(4)}}{n}$$
Its worth to mention that the first two identities were obtained by Cornel and can be found in his paper here using simple integration by parts and clever manipulations, but these three identities and more can be found also in Cornel's book, (Almost) Impossible Integral, Sum, and Series page $59-63$ using only series manipulations.
Also, Ramya showed in his paper here the first three identities using the derivative of $\frac{\partial}{\partial m}\text{B}(n,m)=-\text{B}(n,m)\sum_{k=0}^{n-1}\frac{1}{k+m}$ which we intend here to prove.