求1/(1-x^2)^2的不定积分

Answer 1

关键步骤在于“有理分式的分解”，参考过程如下，望采纳

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图2

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Answer 2

先分式裂：
1/(1-x²)² = {1/2[1/(1+x)+1/(1-x)]}² = (1/4)*1/(1+x)²+(1/4)*1/(1-x)²+(1/2)*1/(1+x)(1-x)]
= (1/4)*1/(1+x)²+(1/4)*1/(1-x)²+(1/4)*1/(1+x)+1/4*(1-x)
∫dx/(1-x²)²
= ∫ [ (1/4)*1/(1+x)²+(1/4)*1/(1-x)²+(1/4)*1/(1+x)+1/4*(1-x) ]dx
= -1/[4(1+x)] + 1/[4(1-x)] + (1/4)ln|1+x| - (1/4)ln|1-x|
= x/[2(1-x²)] + (1/4)ln|(1+x)/(1-x)|

Answer 3

如下图所示，
请认真查看：