∫1/x*( 根号下(1-x)/(1+x))dx

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∫1/x*(根号下(1-x)/(1+x))dx∫1/x*(根号下(1-x)/(1+x))dx∫1/x*(根号下(1-x)/(1+x))dx令√[(1-x)/(1+x)]=u,则:(1-x)/(1+x)

∫1/x*( 根号下(1-x)/(1+x))dx
∫1/x*( 根号下(1-x)/(1+x))dx

∫1/x*( 根号下(1-x)/(1+x))dx
令√[(1-x)/(1+x)]=u,则:(1-x)/(1+x)=u^2,∴1-x=u^2+xu^2,
∴x(1+u^2)=1-u^2,∴x=(1-u^2)/(1+u^2),
∴dx={[(1-u^2)′(1+u^2)-(1-u^2)(1+u^2)′]/(1+u^2)^2}du
  =-4[u/(1+u^2)^2]du.
∴∫(1/x)√[(1-x)/(1+x)]dx
=-4∫[(1+u^2)/(1-u^2)]u[u/(1+u^2)^2]du
=-4∫{u^2/[(1+u^2)(1-u^2)]}du
=2∫[1/(1+u^2)]du-2∫[1/(1-u^2)]du
=2arctanu-∫[1/(1-u)]du-∫[1/(1+u)]du
=2arctan{√[(1-x)/(1+x)]}+ln|1-u|-ln|1+u|+C
=2arctan{√[(1-x)/(1+x)]}+ln|1-√[(1-x)/(1+x)]|
 -ln|1+√[(1-x)/(1+x)]|+C
=2arctan{√[(1-x)/(1+x)]}+ln|√(1+x)-√(1-x)|
 -ln|√(1+x)-√(1-x)|+C.