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求∫[1+ ^3√(1+ x)]/(1 +x)不定积分?

设1+(X+1)^(1/3)=t,则X=(t-1)^3-1,∴dx=3(t-1)^2 dt不定积分I=∫3(t-1)^2/t dt=3∫(t-2+1/t)dt=3[(t^2)/2-2t+lnt]+C再将t=1+(X+1)^(1/3)代入,请自己作一作吧.这类根式函数的积分,通常都用换元积分法来作.多练几道题就可以掌握了.

x^3+1=(x+1)(x^2-x+1) 分解:1/(x^3+1)=1/3*[1/(x+1)-(x-2)/(x^2-x+1)] ∫1/(x+1)dx=ln|x+1|+c ∫(x-2)/(x^2-x+1)dx =1/2*∫(2x-1-3)/(x^2-x+1)dx =1/2*∫(x^2-x+1)'/(x^2-x+1)dx-3/2*∫1/(x^2-x+1)dx =1/2*ln(x^2-x+1)-3/2*∫1/[(x-1/2)^2+3/4]dx =1/2*ln(x^2-x+1)-3/2*2/

令t=x^(1/3),x=t^3,dx=3t^2dt故原式=∫1/(t+1)*3t^2dt =∫3(t^2-1+1)/(t+1)dt =3∫(t-1)dt+3∫1/(t+1)dt =3(t^2/2-t)+3ln(t+1)+C =3(x^2/3)/2-3x^1/3+3ln|x^1/3+1|+C

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 所以∫[1/(1+x^3)]dx =1/3∫(1/(x+1))dx-1/3∫((x-2)/(x^2-x+1))dx 其中1/3∫(1/(x+1))dx=1/3ln|x+1|+c 因为d(x^2-x+1)=(2x-1)dx,所以x-2=1/2(2x-1)-3/2

∫dx/(x2+x+1) =4∫dx/(4x2+4x+1+3) =4∫dx/[(2x+1)2+3] = 4/3∫dx/{[(2x+1)/√3]2+1} = 2/√3∫d[(2x+1)/√3]/{[(2x+1)/√3]2+1} =2arctan[(2x+1)/√3]/√3+C

令(x+1)^1/3=t,x=t^3-1,dx=3t^2dt∫dx/[1+(x+1)^1/3]=∫3t^2/(1+t)dt=3∫t^2/(1+t)dt=3∫(t^2-1+1)/(1+t)dt=3∫[t-1+1/(1+t)]dt=3/2t^2-3t+3ln(1+t)+C反带入即可=3/2(x+1)^2/3-3(x+1)^1/3+3ln(1+(x+1)^1/3)+C扩展资料不定积分的公式1、∫ a dx = ax + C,a和C都

∫√x(1+3√x)dx=∫x^(1/2)*(1+x^(1/3))dx=∫[x^(1/2)+x^(5/6)]dx=∫x^(1/2)dx+∫x^(5/6)dx=x^(1+1/2)/(1+1/2)+x^(1+5/6)/(1+5/6)+C=2/3*x^(3/2)+6/11*x^(11/6)+C

t=√(x+1),dx=2tdt∫1/[1+(√(x+1))^3]dx=∫2tdt/(1+t^3)由于2t/(1+t^3)=(2/3)1/(1+t)+(2/3)(2t-t)/(t^2-t+1)+1/(t^2-t+1)所以:∫1/[1+(√(x+1))^3]dx=∫2tdt/(1+t^3)=∫(2/3)1/(1+t)dt+(1/3)∫(2t-t)/(t^2-t+1)dt+∫1/(t^2-t+1)dt=(2/3)ln(1+t)+(1/3)ln(t^2-t+1)+(2/√3)arctan((2t

显然[1+√(1+x)] *[1-√(1+x)]=1 -1- x= -x于是得到∫x/[1+√(1+x)]dx=∫ -1+ √(1+x) dx代入基本公式∫x^n dx=1/(n+1) *x^(n+1)原积分= -x +2/3 *(1+x)^(3/2) +C,C为常数

用换元法令(x+1)^(1/6)=t√(x+1)=t^3x=t^6-1dx=6t^5dt∫((1-√(x+1))/(1+(x+1)^(1/3)))dx=∫(1-t^3)/(1+t^2)*6t^5dt=6∫(t^5-t^8)/(1+t^2)dt=6∫t^5/(1+t^2)dt-6∫t^8/(1+t^2)dt=3∫t^4/(1+t^2)dt^2-6∫(t^8-1+1)/(1+t^2)dt=3∫(t^4-1+1)/(1+t^2)dt^2-6∫(t^8-1+1)/(1

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