求不定积分∫{√[(1+x)/(1-x)]}/xdx

Question

求不定积分
∫{√[(1+x)/(1-x)]}/xdx

Answer 1

(1+x)/(1-x)>=0
1>x>=-1
令x=cos2A
∫{√[(1+x)/(1-x)]}/xdx
=∫cosA/(sinAcos2A) dcos2A
=∫-2sin2AcosA/(sinAcos2A) dA
=∫-4cosA^2/cos2A dA
=∫-4cosA^2/(2cos2A^2-1) dA
=∫-4cosA^2/(cosA^2-sinA^2) dA
=∫-4/(1-tanA^2) dA (令y=tanA)
=∫-4/[(1-y^2)(1+y^2)]dy
=∫-4/(1-y^4)dy
=∫-1/(1-y)dy +∫-1/(1+y)dy +∫-2/(1+y^2)dy
=ln(1-y)-ln(1+y)-2arctany+C
=ln[(1-tanA)/(1+tanA)]-2A+C
(cosA=[(x+1)/2]^(1/2)sinA=[(1-x)/2]^(1/2) )
=ln[(1-(1-x^2)^(1/2)]-arccosx+C
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