如题所述
let
u=√(x+1)
du = dx/[2√(x+1)]
dx = 2u du
∫[√ (x+1)-1]/[√(x+1)+1 ] dx
=∫[(u-1)/(u+1)] (2udu)
=∫[1 - 2/(u+1)] (2udu)
=∫ [ 2u - 4u/(u+1) ] du
=∫ [ 2u - 4 + 4/(u+1) ] du
=u^2 -4u +4ln|u+1| +C
=(x+1) -4√(x+1) + 4ln|√(x+1)+1| +C