lnx⼀2dx的不定积分怎么求

2025-04-02 18:01:59
推荐回答(2个)
回答1:

∫ln(x/2)dx
=2∫ln(x/2)d(x/2)
=2(x/2)ln(x/2)-2∫x/2dln(x/2)
=x·ln(x/2)-2∫(x/2)·1/xdx
=x·ln(x/2)-∫dx
=xln(x/2)-x+C

回答2:

∫ln(x/2)dx==x·ln(x/2)-∫xdln(x/2)=x·ln(x/2)-∫(x*(2/x))dx=x·ln(x/2)-∫2dx=xln(x/2)-2x+C