根号下x2+1的不定积分是多少

Question

根号下x2+1的不定积分是多少

Answer 1

L=∫√(x^2+1) dx
设x=tanz,dx=(secz)^2 dz
L=∫(secz) * (secz)^2 dz=∫secz d(tanz)
=secz * tanz - ∫tanz d(secz)
=x√(x^2+1) - ∫tanz * (secztanz) dz
=x√(x^2+1) - ∫((secz)^2 - 1) * secz dz
=x√(x^2+1) - L + ∫secz dz
2L = x√(x^2+1) + ∫secz dz
L = (1/2)x√(x^2+1) + ln|secz+tanz| + C
= (1/2)x√(x^2+1) + ln|x + √(x^2+1)| + C