求不定积分∫(x^2/(1+x^4))dx
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解决时间 2021-11-12 13:20
- 提问者网友：箛茗
- 2021-11-11 16:55
求不定积分∫(x^2/(1+x^4))dx
最佳答案
- 五星知识达人网友：爱难随人意
- 2021-11-11 17:40
令x=tany
∫(x^2/(1+x^4))dx
=∫(tany^2/(1+tany^4))*(1/(cosy)^2)dy
=∫(siny)^2/((siny)^4+(cosy)^4) dy
=∫(1/2)(1-cos2y)/(1-4(siny)^2(cosy)^2) dy
=(1/2)∫(1-cos2y)/(1-(sin2y)^2) dy
=(1/2)∫1/(1-(sin2y)^2) dy - (1/2)∫cos2y/(1-(sin2y)^2) dy
=(1/4)∫(1/(cos2y)^2)d(2y) - (1/4)∫1/((1-sin2y)(1+sin2y)) d(sin2y)
=(1/4)tan2y - (1/8)∫(1/(1-sin2y) + 1/(1+sin2y))d(sin2y)
=(1/4)tan2y - (1/8)ln((1+sin2y)/(1-sin2y)) + C
=(1/4)tan2y - (1/4)ln|(siny+cosy)/(siny-cosy)| + C
=(1/2)tany/(1-(tany)^2) - (1/4)ln|(tany+1)/(tany-1)| + C
=(1/2)x/(1-x^2) - (1/4)ln|(x+1)/(x-1)| + C
∫(x^2/(1+x^4))dx
=∫(tany^2/(1+tany^4))*(1/(cosy)^2)dy
=∫(siny)^2/((siny)^4+(cosy)^4) dy
=∫(1/2)(1-cos2y)/(1-4(siny)^2(cosy)^2) dy
=(1/2)∫(1-cos2y)/(1-(sin2y)^2) dy
=(1/2)∫1/(1-(sin2y)^2) dy - (1/2)∫cos2y/(1-(sin2y)^2) dy
=(1/4)∫(1/(cos2y)^2)d(2y) - (1/4)∫1/((1-sin2y)(1+sin2y)) d(sin2y)
=(1/4)tan2y - (1/8)∫(1/(1-sin2y) + 1/(1+sin2y))d(sin2y)
=(1/4)tan2y - (1/8)ln((1+sin2y)/(1-sin2y)) + C
=(1/4)tan2y - (1/4)ln|(siny+cosy)/(siny-cosy)| + C
=(1/2)tany/(1-(tany)^2) - (1/4)ln|(tany+1)/(tany-1)| + C
=(1/2)x/(1-x^2) - (1/4)ln|(x+1)/(x-1)| + C
全部回答
- 1楼网友：神鬼未生
- 2021-11-11 17:50
好像错了哦
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