Question

请问这道不定积分怎么做？谢谢

Answer 1

Answer 2

令 √[(x+1)/x] = u, 则 (x+1)/轿尘x = u^2, x(u^2-1) = 1, x = 1/(u^2-1)
dx = -2udu/(u^2-1)^2
I = ∫-2u^2du/(u^2-1)^2 = (1/2)∫茄帆此[1/(u+1)-1/(u-1)-1/(u+1)^2-1/颤迅(u-1)^2]du
= (1/2)[ln|(u+1)/(u-1)| + 1/(u+1) + 1/(u-1)] + C
= ln[√(x+1)+√x] + (1/2)√x[√(x+1)-√x] + (1/2)√x[√(x+1)+√x] + C

Answer 3

令t=根号((x+1)/x)
t^2 = 1+1/x
x= 1/(t^2-1)
dx = -2t/(t^2-1) dt带人得到
原积分瞎老键=t * -2t/(t^2-1) dt
= -2 -2/(t^2-1) dt
= -2 -1/含指(t-1) +1/(t+1) dt
=-2t +ln((t+1)/(t-1)) +C
然后把t带进去磨巧即可

Answer 4

试试换元法 x=sinh(θ)^2