On the Value Of an Integral

I would like to show you a proof for the integral, $\int_0^1x^{b-1}(1-x)^{c-b-1}\ln (x)dx$ It is for the proving of my previous discussion,"On the Summation of Hypergeometric series". Observe, $\frac{d}{db}(\Beta(b,c-b))=\int_0^1x^{b-1}(1-x)^{c-b-1}(\ln (x)-\ln (1-x))dx$ Then by splitting the integral we get, $\frac{d}{db}(\Beta(b,c-b))=\int_0^1x^{b-1}(1-x)^{c-b-1}\ln (x)dx-\int_0^1x^{b-1}(1-x)^{c-b-1}\ln (1-x)dx$ We have our integral but with an additional integral,for that lets use this integral.Since, $\Beta(b,c-b)=\int_0^1x^{b-1}(1-x)^{c-b-1}dx$ Then taking d.w.r.c,we get, $\frac{d}{dc}(\Beta(b,c-b))=\int_0^1x^{b-1}(1-x)^{c-b-1}\ln (1-x)dx$ And hey! That's our required equation.So, $\frac{d}{db}(\Beta(b,c-b))+\frac{d}{dc}(\Beta(b,c-b))=\int_0^1x^{b-1}(1-x)^{c-b-1}\ln (x)dx$ So the only thing left is to find the value of the derivative,which is quite easy once you have the knowledge of beta function and the diagamma function, $\frac{d}{db}(\Beta(b,c-b))=\Beta(b,c-b)(\psi(b)-\psi(c-b)),\frac{d}{dc}(\Beta(b,c-b))=\Beta(b,c-b)(\psi(c-b)-\psi(c))$ Then adding both derivative we have, $\int_0^1x^{b-1}(1-x)^{c-b-1}\ln (x)dx=\Beta(b,c-b)(\psi(b)-\psi(c))$ And as our corollary ,(I love corollary), $\int_0^1x^{b-1}(1-x)^{-b}\ln (x)dx=\frac{\pi}{\sin(\pi b)}(\psi(b)+\gamma)$ Where $\gamma$ is Euler constant about 0.577.