Math Challenges

Solution:
\( \displaylines{u=\ln\left(x\right),dv=\differentialD x\\ du=\frac{1}{x}\differentialD x,v=x\\ \\ \int\ln\left(x\right)\differentialD x=x\ln\left(x\right)-\int\frac{1}{x}x\differentialD x=x\ln x-x+C} \)