The value of the integral ∫01 xc−1log⁡xdx, c > 0 is

Note that the integral is not an elementary function.Let I(c)=∫01 xc−1log⁡xdx soI′(c)=∫01 xclog⁡xdxlog⁡x=∫01 xcdx=1c+1.Hence I(c)=∫1c+1dc+C=log⁡|c+1|+C.But I(0)=0, hence C=0 so I(c)=log⁡(c+1)