第1个回答 2020-08-26
第2个回答 2020-08-26
令 √[(x+1)/x] = u, 则 (x+1)/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)∫[1/(u+1)-1/(u-1)-1/(u+1)^2-1/(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本回答被提问者采纳
dx = -2udu/(u^2-1)^2
I = ∫-2u^2du/(u^2-1)^2 = (1/2)∫[1/(u+1)-1/(u-1)-1/(u+1)^2-1/(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本回答被提问者采纳
第3个回答 2020-08-26
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=t * -2t/(t^2-1) dt
= -2 -2/(t^2-1) dt
= -2 -1/(t-1) +1/(t+1) dt
=-2t +ln((t+1)/(t-1)) +C
然后把t带进去即可
t^2 = 1+1/x
x= 1/(t^2-1)
dx = -2t/(t^2-1) dt带人得到
原积分=t * -2t/(t^2-1) dt
= -2 -2/(t^2-1) dt
= -2 -1/(t-1) +1/(t+1) dt
=-2t +ln((t+1)/(t-1)) +C
然后把t带进去即可
第4个回答 2020-08-26
试试换元法 x=sinh(θ)^2
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