作业帮x^3/(x+1)^8的不定积分,
来源：学生作业帮助网 编辑：作业帮 时间：2024/04/30 03:29:25
x^3/(x+1)^8的不定积分,
x^3/(x+1)^8的不定积分,
x^3/(x+1)^8的不定积分,
由题意可得:
∫x^3/(x+1)^8dx=∫(x^3+1-1)/(x+1)^8dx
=∫[(x+1)(x^2-x+1)/(x+1)^8-1/(x+1)^8]dx
=∫(x^2-x+1)/(x+1)^7dx-∫1/(x+1)^8dx
=∫(x^2+2x+1-3x)/(x+1)^7dx-∫1/(x+1)^8dx
=∫(x+1)^2/(x+1)^7dx-∫3x/(x+1)^7dx-∫1/(x+1)^8dx
=∫1/(x+1)^5dx-∫(3x+3-3)/(x+1)^7dx-∫1/(x+1)^8dx
=∫1/(x+1)^5dx-∫3(x+1)/(x+1)^7dx+∫3/(x+1)^7dx-∫1/(x+1)^8dx
=∫1/(x+1)^5dx-3∫1/(x+1)^6dx+3∫1/(x+1)^7dx-∫1/(x+1)^8dx
=∫1/(x+1)^5d(x+1)-3∫1/(x+1)^6d(x+1)
+3∫1/(x+1)^7d(x+1)-∫1/(x+1)^8d(x+1)
=-1/4(x+1)^(-4)+3/5(x+1)^(-5)-1/2(x+1)^(-6)+1/7(x+1)^(-7)+C
a=(1+x)^8
x=a^(1/8)-1
dx=1/8*a^(-7/8)da
x^3=a^(3/8)-3a^(1/4)+3a^(1/8)-1
原式=∫[a^(3/8)-3a^(1/4)+3a^(1/8)-1][1/8*a^(-7/8)]/a da
=1/8∫[a^(-3/2)-3a^(-13/8)+3a^(-7/4)-a^(-15/8)]da
=-1/4*a^(-1/2)+3/5*a^(-5/8)-1/2*a^(-3/4)+3/7*a^(-7/8)+C
=-1/4*(1+x)^(-4)+3/5*(1+x)^(-5)-1/2*(1+x)^(-6)+3/7*(1+x)^(-7)+C