求不定积分1.∫[(1/x)√(x–1)]dx令√(x–1)=u,则x-1=u²,x=u²+1；dx=2udu；代入原式得：原式=2∫u²du/(u²+1)=2∫[1-1/(u²+1)]du=2(u-arctanu)+C=2[√(x–1)-arctan√(x–1)+C2.∫x|x–1|dx...