E 1+ ∫LN XDX X LN X= −∫DX X(LN X 1) C= − + => S= −X(LN X 1)− 1E+X(LN X 1)− 1E =2(1−1E)

1/ - DE THI CO DAP AN 12 1/ - DE THI CO DAP AN 12
DE THI CO DAP AN 12
1/e 1+

∫

ln xdx x ln x= −

∫

dx x(ln x 1) C= − + => S= −x(ln x 1)− 11/e+x(ln x 1)− 1e =2(1−1e)