求不定积分 ∫ (1-Inx)⼀(x-Inx)^2 dx

2024-12-28 03:58:26
推荐回答(3个)
回答1:

采用分部积分法,详细看图:



回答2:

解:
∫(1-Inx)/(x-Inx)^2 dx
= ∫(1-Inx)/[x²(1-Inx/x)²] dx
= ∫[1/(1-Inx/x)²]*(1-Inx)/x²dx
= ∫[1/(1-Inx/x)²]d(lnx/x)
= -∫[1/(1-Inx/x)²d(1-lnx/x)
=1/(1-Inx/x) + c =x/(x-lnx) + c

回答3:

答案是:-x/(x-Inx)+C