原式=∫1/√[(x+1)²+4]d(x+1)
設(shè)x+1=2tant,t=actan[(x+1)/2],
則√[(x+1)²+4]=√[4(tan²t+1)]=√(4sec²t)=2sect,d(x+1)=2sec²tdt
∴原式=∫1/√[(x+1)²+4]d(x+1)
=∫1/(2sect)*2sec²tdt
=∫sectdt
=ln|sect+tant|+C
=ln|sec(actan[(x+1)/2])+[(x+1)/2]|+C
=ln|√(1+[(x+1)/2]²)+[(x+1)/2])|+C