求不定积分 ∫8/x^2-4 dx

求不定积分求1/(1+x^3)的不定积

求1/(1+x^3)的不定积分 利用立方和公式a^3 +b^3=(a+b)(a^2-ab+b^2) 1+x^3=(x+1)(x^2-x+1) 1/(1+x^3)=1/(x+1)(x^2-x+1) =1/3[1/(x+1)-(x-2)/(x^2-x+1)] ∫1/(1+x^3)dx=∫1/3[1/(x+1)-(x-2)/(x^2-x+1)]dx =∫1/3{1/(x+1)-[(x-1/2)+3/2]/[(x-1/2)^2+(√3/2)^2]}dx =∫1/3{1/(x+1)-[(x-1/2)+3/2]/[(x-1/2)^2+(√3/2)^2]dx =∫1/3{1/(x+1)-(x-1/2)/...全部

  求1/(1+x^3)的不定积分 利用立方和公式a^3 +b^3=(a+b)(a^2-ab+b^2) 1+x^3=(x+1)(x^2-x+1) 1/(1+x^3)=1/(x+1)(x^2-x+1) =1/3[1/(x+1)-(x-2)/(x^2-x+1)] ∫1/(1+x^3)dx=∫1/3[1/(x+1)-(x-2)/(x^2-x+1)]dx =∫1/3{1/(x+1)-[(x-1/2)+3/2]/[(x-1/2)^2+(√3/2)^2]}dx =∫1/3{1/(x+1)-[(x-1/2)+3/2]/[(x-1/2)^2+(√3/2)^2]dx =∫1/3{1/(x+1)-(x-1/2)/[(x-1/2)^2+(√3/2)^2]+3/2/[(x-1/2)^2+(√3/2)^2]}dx =1/3∫1/(x+1)dx-1/3∫(x-1/2)/[(x-1/2)^2+(√3/2)^2]dx+ 1/2∫/[(x-1/2)^2+(√3/2)^2]dx =1/3ln|x+1|-1/6ln|[(x-1/2)^2+(√3/2)^2]|+1/2*1/(√3/2)arctan[(x-1/2)/(√3/2)+C =1/3ln|x+1|-1/6ln|[(x-1/2)^2+(√3/2)^2]|+4/(3√3)*arctan[(x-1/2)/(√3/2)+C 。
  收起