x^3/(x+1)^8的不定积分,
2个回答
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
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