Evaluate ∫2xx2+1x2+2dx
logx2+1−logx2+2+C
logx2+2−logx2+1+C
logx2+1+logx2+2+C
None of these
Let I=∫2xx2+1x2+2dx
Putting x2=t and 2xdx=dt, we get
I=∫dt(t+1)(t+2)=∫1t+1−1t+2dt=log|t+1|−log|t+2|+C=logx2+1−logx2+2+C