第2个回答 2011-12-02
∫√[(1-x)/(1+x)]dx/x
x=cosu dx=-sinudu sinu=√(1-x^2) u=arccosu
=∫√[(1-cosu)/(1+cosu)]dcosu/cosu
=∫[(1-cosu)/sinu](-sinu)du/cosu
=∫(1-cosu)du/cosu
=∫du/cosu-u
=∫dsinu/[(1-sinu)(1+sinu)] -u
=(1/2)ln|(1+sinu)/(1-sinu)-u+C
=ln|(1+sinu)/(cosu) -u+C
=ln|(1+√(1-x^2))/x| -arccosx+C本回答被提问者采纳