求 ∫√((x+1)/(x-1))+√((x-1)/(x+1))dx
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求 ∫√((x+1)/(x-1))+√((x-1)/(x+1))dx
求 ∫√((x+1)/(x-1))+√((x-1)/(x+1))dx
求 ∫√((x+1)/(x-1))+√((x-1)/(x+1))dx
∫√{[(x+1)/(x-1)]+√[(x-1)/(x+1)]}dx
(x+1)/(x-1)＞0
x＞1或x＜-1
-1＜1/x＜1
设1/x=cosβ ,0＜β＜π
∫√{[(x+1)/(x-1)]+√[(x-1)/(x+1)]}dx
=∫√{[(1+1/x)/(1-1/x)]+√[(1-1/x)/(1+1/x)]}dx
=∫√{[(1+cosβ)/(1-cosβ)]+√[(1-cosβ)/(1+cosβ)]}d(1/cosβ)
=∫[cot(β/2)+tan(β/2)]d(1/cosβ)
=∫[cot(β/2)+tan(β/2)]d(1/cosβ)
=∫[cos(β/2)/sin(β/2)+sin(β/2)/cos(β/2)]d(1/cosβ)
=∫{1/[cos(β/2)sin(β/2)]}d(1/cosβ)
=2∫(1/sinβ)d(1/cosβ)
=-2∫[(1/sinβ)(cosβ)^(-2)]d(cosβ)
=2∫[(1/sinβ)(cosβ)^(-2)]sinβdβ
=2∫[(cosβ)^(-2)]dβ
=2tanβ+c
=2tanarccos(1/x)+c
=2sinarccos(1/x)/cosarccos(1/x)+c
=2xsinarccos(1/x)+c
=2x√{1-[cosarccos(1/x)]^2}+c
=2x√(1-1/x^2)+c
=2(x/|x|)√(x^2-1)+c
求 ∫{√[(x+1)/(x-1)]+√[(x-1)/(x+1)]}dx
解∫{√[(x+1)/(x-1)]+√[(x-1)/(x+1)]}dx =∫√[(x+1)/(x-1)]dx+∫√[(x-1)/(x+1)]dx
令(x+1)/(x-1)=u²；x=(u²+1)/(u²-1)；dx=-[4u/(u²-1)²]du；
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求 ∫{√[(x+1)/(x-1)]+√[(x-1)/(x+1)]}dx
解∫{√[(x+1)/(x-1)]+√[(x-1)/(x+1)]}dx =∫√[(x+1)/(x-1)]dx+∫√[(x-1)/(x+1)]dx
令(x+1)/(x-1)=u²；x=(u²+1)/(u²-1)；dx=-[4u/(u²-1)²]du；
故∫√[(x+1)/(x-1)]dx=∫[-4u²/(u²-1)²]du=-∫[1/(u-1)+1/(u+1)]²du
=-∫[1/(u-1)²+1/(u+1)²+2/(u-1)(u+1)]du
=-∫[1/(u-1)²+1/(u+1)²+1/(u-1)-1/(u+1)]du
=-∫d(u-1)/(u-1)²-∫d(u+1)/(u+1)²-∫d(u-1)/(u-1)+∫d(u+1)/(u+1)
=1/(u-1)+1/(u+1)-ln(u-1)+ln(u+1)+c
=2u/(u²-1)+ln[(u+1)/(u-1)]+c
=(x-1)√[(x+1)/(x-1)]+ln{√[(x+1)/(x-1)]+1}/{√[(x+1)/(x-1)]-1}+c
=√(x²-1)+ln{[√(x+1)+√(x-1)]/[√(x+1)-√(x-1)]+c
=√(x²-1)+ln[x+√(x²-1)]+c
第二个积分∫√[(x-1)/(x+1)]dx也可照此办理，请自己作。
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