求∫x/(1 - x^2)^(1/2)dx,最好有过程

∫x/(1-x²)^(1/2)dx=(1/2)∫ 1/(1-x²)^(1/2)d(x²)=-(1/2)∫ (1-x²)^(-1/2)d(1-x²)=-(1-x²)^(1/2)+C

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

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.

1/x^2(1+x^2)不定积分

∫1/x^2(1+x^2)dx=∫[1/x^2-1/(1+x^2)]dx=-1/x-arctanx+C

求不定积分.∫dx/(x^2 - x - 2) ∫dx/(x^2+x+2) ∫(x+1)dx/(x^2+1) 紧.

∫ dx/(x²-x-2)= ∫ dx/[(x-2)(x+1)]= 1/3*∫ dx/(x-2) - 1/3*∫ dx/(x+1)= 1/3*ln|x-2| - 1/3*ln|x+1| + C= (1/3)ln|(x-2)/(x+1)| + C∫ dx/(x²+x+2)= ∫ dx/[(x+1/2)²+7/4]= (2/√7)arctan[(2x+1)/√7] + C∫ (x+1)/(x²+1) dx= ∫ x/(x²+1) dx + ∫ dx/(x²+1)= (1/2)∫ d(x²+1)/(x²+1) + ∫ dx/(x²+1)= (1/2)ln(x²+1) + arctan(x) + C

求(x - 1)/[(x+1)(x^2+1)]的不定积分

∫(x-1)/[(x+1)(x^2+1)]dx=∫[x/(x^2+1)-1/(x+1)]dx=1/2*log(x^2+1)-log(x+1)+C

求不定积分 ∫x^2/1+x^2 dx 要有详细过程,谢谢

∫x^2+1-1/1+x^2dx=∫1dx-∫1/1+x^2dx=x-arctanx+C

∫x^2/(1+x^2)dx=?

∫ (x²+1-1)/(x²+1)dx=∫ (1 -1/(x²+1) )dx=x-arctanx+C

1/((x^2)(1+x^2))的不定积分

∫dx/x^2(1+x^2)=∫dx[1/x^2-1/(1+x^2)]=∫dx/1/x^2-∫dx/(1+x^2)=-1/x-arctanx+c

∫x^2+1/(x+1)^2(x - 1)dx

解: ∫(x-1)dx/(x²+1)=∫[x/(x²+1)-1/(x²+1) ] dx=∫x/(x²+1) dx-∫1/(x²+1) dx=½∫ 1/(x²+1) d(x²+1)-∫1/(x²+1) dx=½ln(x²+1)-arctanx+c

求(x(x)^1/2)^1/2(x^2 - x^ - 2)dx的积分

题目不清楚