求不定积分,∫x cosx dx; ∫x e(-x)dx; (-x)为幂函数
人气:189 ℃ 时间:2020-04-07 15:55:05
解答
∫ xcosxdx
=∫ x d(sinx)
=xsinx - ∫ sinxdx
=xsinx + cosx + C
∫ xe^(-x)dx
= -∫ x e^(-x)d(-x)
= -∫ x d[e^(-x)]
= - {x[e^(-x)] - ∫[e^(-x)]dx}
= - {x[e^(-x)] + e^(-x)+C1}
= -x[e^(-x)] - e^(-x)+C
推荐
猜你喜欢
- He felt ---at the news that he failed the exam A.frustrating B.rustration C.rustrated D.rustrate
- 如何画英语脑图
- 用所给词的适当形式填空,1,Tom's (.result)of the English exam is pretty good.
- 孩子挤出来(改成比喻句)
- 2x的平方-3y-2(x的平方-2y)
- 一道简单的物理题(质点运动)
- 甲,乙两地相距162千米,一列慢车从甲站出发,每小时走48千米,一列快车从乙站开出,每小时走60千米,试问:1.两列火车同时相向而行,多长时间可以相遇?2.两车同时反向而行,几小时后两车相距270千米?3.若两车相向而行,慢车先开出1小时,
- Look!That woman looks like our teacher .we are going to put it on the day after tomorrow.