$$\int_0^\infty \frac{x^2\ln{x}}{e^x-1}\,dx=\int_0^\infty x^2 e^{-x}\ln x\sum_{k=0}^\infty e^{-kx}$$ $$=\sum_{k=0}^\infty\frac{3-2\gamma-2 \ln (k+1)}{(k+1)^3}$$ $$=(3-2\gamma)\zeta(3)+2\zeta^\prime(3).$$ The integral over $x^2 e^{-(k+1)x}\ln x$ follows upon partial integration and in the final equation I used the identity $$\sum_{k=1}^\infty k^{-p}\ln k=-\zeta'(p).$$

I don't have a published source for this integral, but if need be you could refer to the following derivation: $$\int_0^\infty \frac{x^2\ln{x}}{e^x-1}\,dx=\int_0^\infty x^2 e^{-x}\ln x\sum_{k=0}^\infty e^{-kx}$$ $$=\sum_{k=0}^\infty\frac{3-2\gamma-2 \ln (k+1)}{(k+1)^3}$$ $$=(3-2\gamma)\zeta(3)+2\zeta^\prime(3).$$ The integral over $x^2 e^{-(k+1)x}\ln x$ follows upon partial integration and in the final equation I used the identity $$\sum_{k=1}^\infty k^{-p}\ln k=-\zeta'(p).$$

$$\int_0^\infty \frac{x^2\ln{x}}{e^x-1}\,dx=\int_0^\infty x^2 e^{-x}\ln x\sum_{k=0}^\infty e^{-kx}$$ $$=\sum_{k=0}^\infty\frac{3-2\gamma-2 \ln (k+1)}{(k+1)^3}$$ $$=(3-2\gamma)\zeta(3)+2\zeta^\prime(3).$$ The integral over $x^2 e^{-(k+1)x}\ln x$ follows upon integration by parts.

$$\int_0^\infty \frac{x^2\ln{x}}{e^x-1}\,dx=\int_0^\infty x^2 e^{-x}\ln x\sum_{k=0}^\infty e^{-kx}$$ $$=\sum_{k=0}^\infty\frac{3-2\gamma-2 \ln (k+1)}{(k+1)^3}$$ $$=(3-2\gamma)\zeta(3)+2\zeta^\prime(3).$$ The integral over $x^2 e^{-(k+1)x}\ln x$ follows upon partial integration and in the final equation I used the identity $$\sum_{k=1}^\infty k^{-p}\ln k=-\zeta'(p).$$