Question

求x/[(2-x^2)√(1-x^2)]dx的不定积分，请列出详细解题步骤，谢谢

Answer 1

简单分析一下，答案如图所示

Answer 2

看看这题
解：由于(x^2-x+1)`=2x-1=2(x+1)-3
所以x+1=(1/2)[(2x-1)+3]，故
∫[(x+1)/(x^2-x+1)]dx
=(1/2)∫[(2x-1)+3]dx/(x^2-x+1)
=(1/2)∫(2x-1)dx/(x^2-x+1)+(3/2)∫dx/(x^2-x+1)
=(1/2)∫(x^2-x+1)`dx/(x^2-x+1)+(3/2)∫dx/[(x-1/2)^2+3/4]
=(1/2)∫d(x^2-x+1)/(x^2-x+1)
+(3/2)∫d(x-1/2)/[(x-1/2)^2+(√3/2)^2]
=(1/2)ln|x^2-x+1|+[(3/2)/(√3/2)]arctan[(x-1/2)/(√3/2)]+C
=(1/2)ln(x^2-x+1)+√3arctan[(2x-1)/√3]+C

Answer 3

令x＝sin(t)，dx＝cos(t)dt
原式＝sin(t)cos(t)dt/(2-sin²(t))cos(t)
2-sin²(x)变为1＋cos²(t)，sin(t)放进微分，
有:-dcos(t)/1+cos²(t)
积分等于-arctan(cos(t))+C，
换元变量带回-arctan(cos(arcsin(x)))+C

Answer 4

用分步积分法 ∫x^2 e^(-x)dx=-∫x^2 d(e^(-x))=-x^2 e^(-x)+∫2x e^(-x) dx+c1 =-x^2 e^(-x)-∫2x d(e^(-x))+c1=-x^2 e^(-x)-2x e^(-x) +2∫e^(-x)dx+c2 =-x^2 e^(-x)-2x e^(-x) -2e^(-x)+c3 小宝空间欢迎来踩 <a href="https://wenwen.sogou.com/login/redirect?url=http%3a%2f%2fhi.baidu.com%2f%25cd%25f5%25d0%25a1%25b1%25a6%2f" target="_blank">http://hi.baidu.com/%cd%f5%d0%a1%b1%a6/</a>