∫x−1xx+1dx is equal to
lnx−x2−1−tan−1x+C
lnx+x2−1−tan−1x+C
lnx−x2−1−sec−1x+C
lnx+x2−1−sec−1x+C
I=∫x−1xx+1dx=∫x−1xx2−1=∫dxx2−1−∫dxxx2−1=lnx+x2−1−sec−1x+C