求广义积分∫(上限1,下限-1)dx⼀x(x-2)

2025-02-23 10:31:14
推荐回答(1个)
回答1:

∫ dx/x(x-2)
=1/2 ∫1/(x-2) -1/x dx
=1/2 *ln|(x-2)/x|
代入上下限1和-1
=1/2 *ln1 -1/2 *ln3= -1/2 *ln3