设x-1=√3tant,则dx=√3sec²tdt,sint=(x-1)/√(x²-2x+4),cost=√3/√(x²-2x+4)
∴原式=∫dx/√[3+(x-1)²]
=∫√3sec²tdt/(√3sect)
=∫dt/cost
=∫costdt/cos²t
=∫d(sint)/(1-sin²t)
=1/2∫[1/(1+sint)+1/(1-sint)]d(sint)
=1/2(ln│1+sint│-ln│1-sint│)+C1(C1是积分常数)
=1/2ln│(1+sint)/(1-sint)│+C1
=ln│(1+sint)/cost│+C1
=ln│x-1+√(x²-2x+4)│-1/2ln3+C1
=ln│x-1+√(x²-2x+4)│+C (其中C=C1-1/2ln3).