∫(u+1)/(u²-u+1)du=1/2∫(2u-1+3)/(u²-u+1)du=1/2∫d(u²-u+1)/(u²-u+1)du+3/2∫1/(u²-u+1)du=1/2ln(u²-u+1)+√3∫1/[((2u-1)/√3)²+1]d((2u-1)/√3)=1/2ln(u²-u+1)+√3arctan((2u-1)/√3)+C
