Question

定积分.广义积分.解题不能,

Answer 1

原式=∫(1→2)dx/[(x-1)(x-3)]=1/2∫(1→2)(1/(x-3)-1/(x-1))dx=1/2∫(1→2)dx/(x-3)-1/2∫(1→2)dx/(x-1)=1/2ln|x-3||(1→2)-1/2ln|x-1||(1→2)=0-1/2ln2-1/2lim(t→1+)ln(t-1)+0=+∞