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

简单分肆陆析裂陵顷一下，答案汪启如图所示

Answer 2

看闷笑看这题
解：由于(x^2-x+1)`=2x-1=2(x+1)-3
所以x+1=(1/2)[(2x-1)+3]，氏锋故
∫[(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+(√歼罩晌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)放进微困帆分，
有:-dcos(t)/键冲1+cos²(t)
积分等于-arctan(cos(t))+C，
换元变汪亮雹量带回-arctan(cos(arcsin(x)))+C

Answer 4

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