Presupun ca stim [tex]\int\ln x\ dx=x(\ln x-1)[/tex] .
Atunci integrala ceruta se scrie, folosind integrarea prin parti:
[tex]=\int\ln x\ln x\ dx=\\ \\ =\int[x(\ln x-1)]'\ln x\ dx=\\ \\ \\ =x(\ln x-1)\ln x-\int x(\ln x-1)\cdot \frac{1}{x}=\\ \\ \\ =x(\ln x-1)\ln x-\int(\ln x-1)\ dx=\\ \\ \\ =x(\ln x-1)\ln x-x(\ln x-1)+x+C=\\ \\ \\ =x[(\ln x-1)^2+1]+C[/tex]
