Question

1/根号下[（x-1/2）^2-1/4]的不定积分怎么求

Answer 1

令 x - 1/2 = (1/2)secu, 则 dx = (1/2)secutanudu，
I = ∫dx/√[(x-1/2)^2-1/4] = ∫(1/2)secutanudu/[(1/2)tanu]
= ∫secudu = ln|secu+tanu| + C = ln|2x-1+2√(x^2-x)| + C

Answer 2

∫√(x^2-1)dx
设x=sect,dx=secttantdt
=∫√[(sect)^2-1]*secttantdt
=∫√(tant)^2*secttantdt
=∫(tant)^2*sectdt
=∫(tant)^2*sectdt
=∫((sect)^2-1)*sectdt
=∫sectdt-∫(sect)^3dt
=ln(sect+tant)+
∫sectdtant
=ln(sect+tant)+
secttant-∫tantdsect
=ln(sect+tant)+secttant-∫(tant)^2sectdt
得
∫√(x^2-1)dx
=∫(tant)^2sectdt
=1/2[ln(sect+tant)+
secttant]
由x=sect,得tant=√(x^2-1)
=1/2[ln(x+√(x^2-1))+x√(x^2-1)]

Answer 3

let
x-1/2 = (1/2)secu
dx=(1/2)secu.tanu du
∫dx/√[(x-1/2)^2-1/4]
=∫(1/2)secu.tanu du/[(1/2)tanu]
=∫secu du
=ln|secu+tanu| +C
=ln|(2x-1)+√[(x-1/2)^2-1/4] | +C