Please solve the question with fill solution. Thanks.?
回答 (2)
2016-02-04 6:59 am
let u e^(1/x)
lnu = 1/x
1/u du = -1/x² dx
∫e^(1/x) / x² dx
=－∫ 1 du
= - u + C
= - e^(1/x) + C
lnu = 1/x
1/u du = -1/x² dx
∫e^(1/x) / x² dx
=－∫ 1 du
= - u + C
= - e^(1/x) + C
2016-02-04 3:40 am
Sol
∫e^(1/x)/x^2dx
=－∫e^(1/x)/x^2dx
=－∫e^(1/x)d(1/x)
=－e^(1/x)+c
∫e^(1/x)/x^2dx
=－∫e^(1/x)/x^2dx
=－∫e^(1/x)d(1/x)
=－e^(1/x)+c
收錄日期: 2021-04-18 14:29:42
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