求不定积分求1/(1+x^3)的不定积分
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
。
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