I = ∫dx/[1+√(1+x)]
令 1+√(1+x) = u,则 x=(u-1)^2-1 = u^2-2u,
得 I = ∫(2u-2)du/u = 2∫(1-1/u)du
= 2(u-lnu)+C = 2+2√(1+x)-2ln[1+√(1+x)]]+C
I = ∫dx/[1+√(1+x)]
令 1+√(1+x) = u,则 x=(u-1)^2-1 = u^2-2u,
得 I = ∫(2u-2)du/u = 2∫(1-1/u)du
= 2(u-lnu)+C = 2+2√(1+x)-2ln[1+√(1+x)]]+C