求(x^2+1)/(x^2-2x+2)^2 不定积分
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解决时间 2021-11-11 14:32
- 提问者网友：愿为果
- 2021-11-10 16:46
求(x^2+1)/(x^2-2x+2)^2 不定积分
最佳答案
- 五星知识达人网友：野慌
- 2021-11-10 18:23
(x^2+1)/(x^2-2x+2)^2
=[(x-1）^2+2(x-1)]/[(x-1)^2+1]^2
={[(x-1）^2+1]+2(x-1)-1]}/[(x-1)^2+1]^2
=1/[(x-1)^2+1]+[2(x-1)-1]/[2(x-1)-1]
因此，积分(x^2+1)/(x^2-2x+2)^2dx
=积分1/[(x-1)^2+1]d(x-1)+积分1/[(x-1)^2+1]^2 d[(x-1)^2+1]-积分1/[(x-1)^2+1]^2 dx
=arctan(x-1)-1/[(x-1)^2+1]-(x-1)/{2[(x-1)^2+1]}+(1/2)arctan(x-1)+c
其中求积分1/[(x-1)^2+1]^2 dx时
可设x-1=tany
=[(x-1）^2+2(x-1)]/[(x-1)^2+1]^2
={[(x-1）^2+1]+2(x-1)-1]}/[(x-1)^2+1]^2
=1/[(x-1)^2+1]+[2(x-1)-1]/[2(x-1)-1]
因此，积分(x^2+1)/(x^2-2x+2)^2dx
=积分1/[(x-1)^2+1]d(x-1)+积分1/[(x-1)^2+1]^2 d[(x-1)^2+1]-积分1/[(x-1)^2+1]^2 dx
=arctan(x-1)-1/[(x-1)^2+1]-(x-1)/{2[(x-1)^2+1]}+(1/2)arctan(x-1)+c
其中求积分1/[(x-1)^2+1]^2 dx时
可设x-1=tany
全部回答
- 1楼网友：青尢
- 2021-11-10 20:02
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