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求1/(x^3+1)的原函数

你好!这是有理函数的积分,可以如图分解拆项后,再用凑微分法计算。经济数学团队帮你解答,请及时采纳。谢谢!

这样

x^(-3+1)/(-3+1)=-1/2 *x^(-2)

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∫dx/(x^6+1)=∫(x^2+1)dx/(x^6+1)-∫x^2dx/(x^6+1) ∵(a+b)(a^2-ab+b^2)=a^3+b^3 ∴∫(x^2+1)dx/(x^6+1)=∫dx/(x^4-x^2+1) =1/2∫(x^2+1)/(x^4-x^2+1)-1/2∫(x^2-1)dx/(x^4-x^2+1) =1/2∫(1+x^-2)/(x^2+x^-2-1)-1/2∫(1-x^-2)dx/(...

:依题意:∫x^3/(1+x^2)dx =∫(x^2 * arctanx)dx =1/3*∫arctanxd(x^3)dx =1/3*x^3*arctanx-1/3*∫x^3/(1+x^2)dx =1/3*x^3*arctanx-1/3*∫(x^3+x-x)/(1+x^2)dx =1/3*x^3*arctanx-1/3*∫xdx+1/3*∫x/(1+x^2)dx =1/3*x^3*arctanx-1/6*x^2..

1/(x+x^3)=x/x^2(1+x^2)=x(1/x^2-1/(1+x^2))=1/x-x/(1+x^2) 所以∫1/(x+x^3)dx=∫1/x-x/(1+x^2)dx=ln|x|-1/2∫1/(1+x^2)d(1+x^2)=ln|x|-1/2ln(1+x^2)+C

先分解因式: ∫ 1/(x³ + 1) dx = ∫ 1/[(x + 1)(x² - x + 1)] dx = ∫ A/(x + 1) dx + ∫ (Bx + C)/(x² - x + 1) dx 1 = A(x² - x + 1) + (Bx + C)(x + 1) = Ax² - Ax + A + Bx² + Cx + Bx + C 1 = (A + B)x² +...

解法一: 思路:根据分子分母的关系,直接变形化简求得: I=-∫[x(1-x^2)-x]dx/√(1-x^2) =-∫x(1-x^2)dx/√(1-x^2)+ ∫xdx/√(1-x^2) =-∫x√(1-x^2)dx-(1/2) ∫d(1-x^2)/√(1-x^2) =(1/2) ∫√(1-x^2)d(1-x^2)-√(1-x^2) =(1/3)√(1-x^2)^3-√(1-x^2)+c END ...

解:1+x^3=(x+1)(x^2-x+1) 用待定系数法:A/(x+1)+(Bx+C)/(x^2-x+1)=1/(x+1)(x^2-x+1) 得A=1/3,B=-1/3,C=2/3 原式=∫dx/[(x+1)(x²-x+1)] =∫[(1/3)/(x+1)+(-x/3+2/3)/(x²-x+1)]dx =(1/3)ln│x+1│+(1/6)∫(3+1-2x)/(x²-x+1)dx =(1/3)l...

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