∫1/(x+1)dx积分从1到3

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∫1/(x+1)dx积分从1到3
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∫1/(x+1)dx积分从1到3
∫1/(x+1)dx积分从1到3

∫1/(x+1)dx积分从1到3
原函数为ln(1+x),积分值等于ln(4)-ln(2)=ln2

∫dx/[(x+1)²(x-1)⁴]^(1/3)的不定积分积分
原式=∫dx/{[(x+1)^(2/3)][(x-1)^(4/3)]}
=∫dx/{[(x+1)^(2/3)][(x-1)^(-2/3)](x-1)²}
=∫{[(x-1)/(x+1)]^(2/3)}dx/(x-1)²
=-(1/2)∫{(x+1)/(x-1)]^(-2/3)}d[(x+1)/(x-1)]
=-(1/2)(3)[(x+1)/(x-1)]^(1/3)+C
=-(3/2)[(x+1)/(x-1)]^(1/3)+C