=13∫x−1x2+x+1dx13⎡⎢
⎢
⎢⎣∫x+12(x+12)2+(√3/2)2dx−32∫dx(x+1/2)2+(√3/22)⎤⎥
⎥
⎥⎦
Put x+1/2=t dx=dt
=13[∫tdtt2+(√3/2)2−32∫dtt2+(√3/2)2]
Let put t2=u 2tdt=du
=13[∫du4+3/4−32×2√3tan−1t×2√3]
=13[12log|(x+1/2)2+3/4|−1√3tan−1(2x+1√3)]+c
=16log|x2+x+1|−13√3tan−1(2x+1√3)+c