I=∫(1x2+1)dx(1x−x)√(x−1x)2+2Putting m=x−1x
⇒dm=(1+1x2)dx
⇒I=−∫dmm√m2+2
Let t=√m2+2
⇒dt=mdm√m2+2⇒dtm=dm√m2+2
I=−∫dtt2−2
=−∫dt(t+√2)(t−√2)=−123/2∫[1t−√2−1t+√2]dt
=−123/2[ln∣∣t−√2∣∣−ln∣∣t+√2∣∣]+c
=123/2ln∣∣t+√2∣∣−123/2ln∣∣t−√2∣∣+c
=123/2ln∣∣√m2+2+√2∣∣−123/2ln∣∣√m2+2−√2∣∣+c
m2+2=(x−1x)2+2
=x2+1x2−2+2=x2+1x2
I=123/2ln∣∣∣√x2+1x2+√2∣∣∣−123/2ln∣∣∣√x2+1x2−√2∣∣∣+c