高等数学求解

高数积分求解求解∫1/(1+x&

设1/(1+x^3)=a/(1+x)+(bx+c)/(1-x+x^2) =[a(1-x+x^2)+(bx+c)(1+x)]/(1+x^3) =[(a+c)+(-a+b+c)x+(a+b)x^2]/(1+x^3) 所以a+c=1,-a+b+c=0,a+b=0 a=1/3,b=-1/3,c=2/3 1/(1+x^3)=(1/3)[1/(1+x)-(x-2)/(1-x+x^2)] ∫[1/(1+x)]dx=ln(1+x)+C1 (x-2)/(1-x+x^2)=(x-1/2-3/2)/[(x-1/2)^2+3/4] =(x-1/2)/[(x-1/2)^2+3/4]-(3/2)/(x-1/2)^2+...全部

  设1/(1+x^3)=a/(1+x)+(bx+c)/(1-x+x^2) =[a(1-x+x^2)+(bx+c)(1+x)]/(1+x^3) =[(a+c)+(-a+b+c)x+(a+b)x^2]/(1+x^3) 所以a+c=1,-a+b+c=0,a+b=0 a=1/3,b=-1/3,c=2/3 1/(1+x^3)=(1/3)[1/(1+x)-(x-2)/(1-x+x^2)] ∫[1/(1+x)]dx=ln(1+x)+C1 (x-2)/(1-x+x^2)=(x-1/2-3/2)/[(x-1/2)^2+3/4] =(x-1/2)/[(x-1/2)^2+3/4]-(3/2)/(x-1/2)^2+3/4] ∫(x-2)/(1-x+x^2)dx =∫(x-1/2)/[(x-1/2)^2+3/4]dx-∫(3/2)/(x-1/2)^2+3/4]dx =(1/2)∫d(x-1/2)^2/[(x-1/2)^2+3/4] -2∫d(x-1/2)/[1+(4/3)(x-1/2)^2] =(1/2)ln[(x-1/2)^2+3/4]-(√3)arctan(2/√3)(x-1/2)+C2 原式 =(1/3)[ln(1+x)-(1/2)ln(1-x+x^2)+(√3)arctan(2/√3)(x-1/2)]+C 以上C1,C2,C都是任意常数 方法是正确的，计算可能有错。
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