∫1x−xdx is equal to
2log|x+1|+C
log|x+1|+C
2log|x−1|+C
∫1x−xdx=∫1x(x−1)dx Let x−1=t⇒12x=dtdx⇒dx=2xdt∴1x(x−1)dx=∫1xt2xdt=∫2tdt=2⋅log|t|+C =2log|x−1|+C