∫xx2+2x+2x+1dx is equal to

∫xx2+2x+2x+1dx=∫xx+12+1x+1dx Put x+1=t2dx=2tdt∫t2−12tdtt4+1t=2∫t21−1t2dtt2t2+1t2=2∫1−1t2t2+1t2dt=2∫1−1t2dtt+1t2−(2)2 Put t+1t=k,1−1t2dt=dk=2∫dkk2−(2)2=2122log⁡k−2k+2+C=12log⁡t+1t−2t+1t+2+C=12log⁡t2−2t+1t2+2t+1+C  =12logx+1−2x+1+1x+1+2x+1+1+C