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求不定积分∫5xln(1+x)dx,请帮手解说

2025-03-16 12:04:22

推荐回答（3个）

回答1：

分部积分
∫5xln(1+x)dx
=(5/2)∫ln(1+x)d(x^2)
=(5/2)(x^2)*ln(1+x)-(5/2)∫(x^2)/(1+x)dx
=(5/2)(x^2)*ln(1+x)-(5/2)∫(x^2-1+1)/(1+x)dx
=(5/2)(x^2)*ln(1+x)-(5/2)∫(x-1)dx-(5/2)∫1/(1+x)dx
=(5/2)(x^2)*ln(1+x)-(5/4)x^2+(5/2)x-(5/2)ln|1+x|+C

回答2：

原式子
=（5/2)∫ln(1+x)d(x^2)
=(5/2)x^2ln(1+x)-(5/2)∫x^2dx/(1+x)
=(5/2)x^2ln(1+x)-(5/2)∫[(x+1)(x-1)+1]dx/(1+x)
=(5/2)x^2ln(1+x)-(5/2)∫(x-1)dx-(5/2)∫d(x+1)/(x+1)
=(5/2)x^2ln(1+x)-(5/2)(x^2/2-x)-(5/2)ln(x+1)+c.

回答3：

∫5xln(1+x)dx
=5/2∫ln(1+x)dx^2
=5/2x^2lnx-5/2∫x^2/(1+x)dx
=5/2x^2lnx-5/2∫[x-1+1/(1+x)]dx
=5/2x^2lnx-5/4x^2+5/2x-5/2ln(1+x)+C

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