[微積] ln的積分
0
∫(1-s)/(100-s)ds
1
let u=1-s du=-ds
dv=(100-s)^-1ds v=-ln(100-s)
0 0
-(1-s)ln(100-s) -∫ln(100-s)ds=ln100-∫ln(100-s)ds
1 1
我的問題是後面的部分怎麼積? 我知道∫lnu du=ulnu-u+c
以下是我寫的
a=100-s da=-ds
100
-∫ lna da
99
100
=-(alna-a)
99
=-100ln100+100+99ln99-99
因此-(1-s)ln(100-s) -∫ln(100-s)ds
=ln100+100ln100-100-99ln99+99
=101ln100-99ln99-1
but the answer is 1+99ln0.99
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