I=∫dxx2√(1+x2)3Đặt x=tant,t∈(−π2;0)∪(0;π2)
⇒dx=dtcos2t
√(1+x2)3=√(1+tan2t)3=(√1cos2t)3=1cos3t (do t∈(−π2;0)∪(0;π2))
⇒I=∫dtcos2tsin2tcos2t1cos3t=∫cos3tsin2tdt=∫(1−sin2t)costsin2tdt
⇒I=∫(1sin2t−1)costdt
Đặt u=sint⇒du=costdt
⇒I=∫(1u2−1)du=−1u−u+c
Ta có:
x=tant=sintcost=u√1−u2
⇒x2(1−u2)=u2
⇒u=x√1+x2
Vậy: I=−√1+x2x−x√1+x2+c