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x x三次方分之一的微分

-1/2x+C

∫x^ndx=x^(n+1)/(n+1)+C n=1/3 所以原式=3x^(4/3)/4+C

这题目很烦的:答案是(1/3)ln(x-1)-(1/6)ln(x+x+1)-(2/√3)arctan[(2x+1)/√3]+c 过程: ∫1/(x-1) dx =∫1/(x-1)(x+x+1) dx =∫1/3(x-1)-(x+2)/3(x+x+1)] dx,总之在/后面的都是分母 =1/3*∫1/(x-1) dx-1/3*∫(x+2)/(x+x+1) dx =1/3*∫1/(x-1) d(x-

∫x/(x^3+1)dx将x替换x=(-1/3)(x^2-x+1)+(1/3)(x+1)(x+1)∫x/(x^3+1)dx=∫(-1/3)x/(x+1)dx+∫(1/3)(x+1)/(x^2-x+1)dx=(1/3)(lnIx+1I-x)+(1/6)∫1/(x^2-x+1)d(x^2-x+1)+(√3/3)arctan[(2x-1)/√3]=(1/3)(lnIx+1I-x)

y=x^(1/3) 那么 y'=lim(dx->0) [(x+dx)^(1/3) -x^(1/3)] /dx 注意由立方差公式可以得到(x+dx)^(1/3) -x^(1/3)=(x+dx -x) / [(x+dx)^(2/3) + (x+dx)^(1/3)*x^(1/3) +x^(2/3)]=dx / [(x+dx)^(2/3) + (x+dx)^(1/3)*x^(1/3) +x^(2/3)] 所以 y'=lim(dx->0) 1 / [(x+dx)^(2/3) +

∫x/(x^3+1)dx 将x替换 x=(-1/3)(x^2-x+1)+(1/3)(x+1)(x+1) ∫x/(x^3+1)dx=∫(-1/3)x/(x+1)dx+∫(1/3)(x+1)/(x^2-x+1)dx =(1/3)(lnIx+1I-x)+(1/6)∫1/(x^2-x+1)d(x^2-x+1)+(√3/3)arctan[(2x-1)/√3] =(1/3)(lnIx+1I-x)+(1/6)lnIx^2-x+1I+(√3/3)arctan[(2x-1)/√3]+C 解答完毕,真心不易,请及时采纳呀

答案是3/4倍的X的4/3次方+C

x^2+x^2分之一=(x-x分之一)^2+2=1+2=3

f(x)=x^(1/3)f'(x)=(1/3)x^(-2/3)当x=0时,x^(2/3)=0,取倒数)x^(-2/3)无意义,故f'(0)不存在垂直的x轴的切线即导数为无穷大,在x=0处取得(可画图)垂直于y轴的切线不存在,因为f'(x)不等于0

∫ x^3/√(1-x^2)dx 代换:令x=sint 则原式=∫ (sint)^3dt=∫ sint [1-(cost)^2]dt=-cost+(-cost)^3+c=-√(1-x^2)+√(1-x^2)^3/3+c

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