Question

大学不定积分一题？

见图，谢谢

Answer 1

具体求法，

如图所示

Answer 2

∫In(x-a)/(x-b)dx =∫[(In(x-a)-In(x-b)]dx =∫In(x-a)dx-∫In(x-b)dx =(x-a)[In(x-a)-1]-(x-b)[In(x-b)-1]+C =(x-a)In(x-a)-(x-b)In(x-b)+b-a+C C是常数. 这里说一下, ∫In(x-a)dx=∫In(x-a)(x)'dx =xIn(x-a)-∫{[In(x-a)]'*x}dx =xIn(x-a)-∫[x/(x-a)]dx =xIn(x-a)-∫[1-a/(x-a)]dx =xIn(x-a)-x+∫a/(x-a)dx =xIn(x-a)-x+aIn(x-a)+C =(x-a)[In(x-a)-1]-a+C a,C看成常数, (x-a)[In(x-a)-1]-a+C与(x-a)[In(x-a)-1}+C是等价的. 这是定积分分部积分法.

Answer 3

I = ∫x^2e^(-x)dx = -∫x^2de^(-x) = -x^2e^(-x) + 2∫xe^(-x)dx
= -x^2e^(-x) - 2∫xde^(-x) = -x^2e^(-x) - 2xe^(-x) + 2∫e^(-x)dx
= -x^2e^(-x) - 2xe^(-x) - 2e^(-x) + C = -(x^2+2x+2)e^(-x) + C