Question

帮忙算一下，要过程，谢谢

Answer 1

∫ x. √[(x-1)/(x+1)] dx
=∫ x(x-1)/√(x^2-1) dx
=∫ x^2/√(x^2-1) dx - ∫ x/√(x^2-1) dx
=∫ x^2/√(x^2-1) dx - (1/2)∫ d(x^2-1)/√(x^2-1)
=∫ x^2/√(x^2-1) dx - √(x^2-1)
=(1/2)[ √(x^2-1) +ln|x+ √(x^2-1)/x | ] - √(x^2-1) + C
=(1/2)[ -√(x^2-1) +ln|x+ √(x^2-1)/x | ] + C

let
x= secu
dx = secu. tanu du

∫ x^2/√(x^2-1) dx
=∫ [(secu)^2/tanu] secu. tanu du
=∫ (secu)^3 du
=∫ secu dtanu
= secu.tanu - ∫ (tanu)^2 .secu du
= secu.tanu - ∫ [(secu)^2-1] .secu du
2∫ (secu)^3 du = secu.tanu +∫secu du
∫ (secu)^3 du = (1/2)[ secu.tanu +ln|secu+ tanu| + C'

=>
∫ x^2/√(x^2-1) dx
=(1/2)[ secu.tanu +ln|secu+ tanu| ] + C'
=(1/2)[ √(x^2-1) +ln|x+ √(x^2-1)/x | ] + C'

cosu = 1/x
tanu =

追问

∫ x^2/√(x^2-1) dx =