求不定积分∫x²+1\(x+1)²(x-1)第一步是怎样拆分的?为什么要拆成x-1 x+1 (x+
求不定积分∫x²+1\(x+1)²(x-1)第一步是怎样拆分的?为什么要拆成x-1x+1(x+1)^2?(x+1)^2(x-1)并没有公因式...
求不定积分∫x²+1\(x+1)²(x-1)第一步是怎样拆分的?为什么要拆成x-1 x+1 (x+1)^2?(x+1)^2(x-1)并没有公因式
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2016-04-08
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拆分是为了将原函数求出来,具体过程如下:
∫{(x^2+1)/[(x+1)^2·(x-1)]}dx
=∫{[(x^2+2x+1)-2x]/[(x-1)(x+1)^2]}dx
=∫{[(x+1)^2-2x]/[(x-1)(x+1)^2]}dx
=∫[1/(x-1)]dx-2∫{x/[(x-1)(x+1)^2]}dx
=∫[1/(x-1)]dx-∫[1/(x+1)][1/(x-1)+1/(x+1)]dx
=∫[1/(x-1)]dx-∫[1/(x+1)][1/(x-1)]dx-∫[1/(x+1)^2]dx
=∫[1/(x-1)]dx-(1/2)∫[1/(x-1)-1/(x+1)]dx-∫[1/(x+1)^2]dx
=(1/2)∫[1/(x-1)]dx+(1/2)∫[1/(x+1)]dx-∫[1/(x+1)^2]dx
=(1/2)ln|x-1|+(1/2)ln|x+1|+1/(x+1)+C。
∫{(x^2+1)/[(x+1)^2·(x-1)]}dx
=∫{[(x^2+2x+1)-2x]/[(x-1)(x+1)^2]}dx
=∫{[(x+1)^2-2x]/[(x-1)(x+1)^2]}dx
=∫[1/(x-1)]dx-2∫{x/[(x-1)(x+1)^2]}dx
=∫[1/(x-1)]dx-∫[1/(x+1)][1/(x-1)+1/(x+1)]dx
=∫[1/(x-1)]dx-∫[1/(x+1)][1/(x-1)]dx-∫[1/(x+1)^2]dx
=∫[1/(x-1)]dx-(1/2)∫[1/(x-1)-1/(x+1)]dx-∫[1/(x+1)^2]dx
=(1/2)∫[1/(x-1)]dx+(1/2)∫[1/(x+1)]dx-∫[1/(x+1)^2]dx
=(1/2)ln|x-1|+(1/2)ln|x+1|+1/(x+1)+C。
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就是为了消掉分子
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