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求不定积分∫x^2+1/(x^2-1)(x+1)dx答案是1/2ln(x^2-1)+1/求不定积分∫x^2+1/(x^2-1)(x+1)dx答案是1/2ln(x^2-1)+1/(x+1）步骤：原式=∫(x^2+x-x+1)/(x^2-1)(x+1)dx=∫x(x+1)/(x^2-1)(x+1)dx-∫(x-1)/(x^2-1)(x+1)dx=∫x/(x^2-1)dx-
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求不定积分 ∫x^2+1/(x^2-1)(x+1)dx 答案是 1/2ln(x^2-1)+1/
求不定积分
∫x^2+1/(x^2-1)(x+1)dx
答案是
1/2ln(x^2-1)+1/(x+1）
步骤 ：原式=∫(x^2+x-x+1)/(x^2-1)(x+1)dx
=∫x(x+1)/(x^2-1)(x+1)dx-∫(x-1)/(x^2-1)(x+1)dx
=∫x/(x^2-1)dx-∫1/(x+1)^2dx
=1/2ln(x^2-1)+1/(x+1)
我想问ln里面x^2-1不加绝对值么？？
求不定积分
∫x^2+1/(x^2-1)(x+1)dx
答案是
1/2ln(x^2-1)+1/(x+1）
步骤 ：原式=∫(x^2+x-x+1)/(x^2-1)(x+1)dx
=∫x(x+1)/(x^2-1)(x+1)dx-∫(x-1)/(x^2-1)(x+1)dx
=∫x/(x^2-1)dx-∫1/(x+1)^2dx
=1/2ln(x^2-1)+1/(x+1)
我想问ln里面x^2-1不加绝对值么？？
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