求不定积分∫1/xdx 不定积分1-x/xdx？
1/1+xdx\u7684\u4e0d\u5b9a\u79ef\u5206\u7528\u7b2c\u4e00\u7c7b\u6362\u5143\u6cd5\u6c42
\u8fc7\u7a0b\u89c1\u4e0a\u56fe
\u5217\u9879\uff0c\u518d\u4e0d\u5b9a\u79ef\u5206\uff0c\u6b65\u9aa4\u5982\u4e0b\uff1a
\u222b(1-x)dx/x
=\u222bdx/x-\u222bdx
=lnx-x+c
因为积分函数y=f(x)=1/x是反比例函数，存在两支
所以：x<0和x>0都要考虑
x>0时积分得：lnx+C
x<0时：
∫ 1/x dx=∫ 1/(-x) d(-x)=ln(-x)+C
综上所述，∫1/x dx=ln|x|+C
x<0时，ln(-x)的导数也是1/x
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