Question

求不定积分∫(1/x)√[(1-x)/(1+x)]dx

Answer 1

解，令t=根号{(1-x)/(1+x)}，则x=(t^2+1)/(t^2-1)代入得
∫1/[(1-x)*(1-x^2)^0.5]dx
=1/2∫（t^2-1）/(t^2+1)dt^2
=1/2[t^2-2ln(t^2+1)]c，其中c为常数
求得解为根号1/2{(1-x)/(1+x)+2ln(1+x)-2ln2}+c

Answer 2

∫√[(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