![](https://iknow-base.cdn.bcebos.com/lxb/notice.png)
不定积分 1/x^4√(1+x^2) dx
2个回答
展开全部
∫ 1/[x^4 * √(1+x²)] dx
Let x = tany and dx = sec²y dy,then √(1+x²) = secy
==> ∫ sec²y / (tan^4y * secy) dy
= ∫ secy * cos^4y / sin^4y dy
= ∫ cscy * cot³y dy
= ∫ cscy * coty * (csc²y - 1) dy
= -∫ (csc²y - 1) dcscy
= cscy - csc³y / 3 + C
= [(2x²-1)√(1+x²)] / (3x³) + C
楼上的答案不对,-1/(3x³) + 1/x - arctan(1/x)的导数是1/(x^4 + x^6) = 1/[x^4 * (1+x²)]
漏了个根号
Let x = tany and dx = sec²y dy,then √(1+x²) = secy
==> ∫ sec²y / (tan^4y * secy) dy
= ∫ secy * cos^4y / sin^4y dy
= ∫ cscy * cot³y dy
= ∫ cscy * coty * (csc²y - 1) dy
= -∫ (csc²y - 1) dcscy
= cscy - csc³y / 3 + C
= [(2x²-1)√(1+x²)] / (3x³) + C
楼上的答案不对,-1/(3x³) + 1/x - arctan(1/x)的导数是1/(x^4 + x^6) = 1/[x^4 * (1+x²)]
漏了个根号
推荐律师服务:
若未解决您的问题,请您详细描述您的问题,通过百度律临进行免费专业咨询