求1/1+根号x+1dx的不定积分

Question

如题所述

Answer 1

let

y=√(x+1)
dy = (1/[2√(x+1)]) dx
dx = 2ydy

∫ dx/[1+√(x+1)]
=∫ [2y/(1+y)] dy
=2∫ dy - 2∫ dy/(1+y)
=2y -2ln|1+y| + C
=2√(x+1) - 2ln|1+√(x+1)| + C