∫（x+1）²dx求此不定积分

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摘要 A=∫x√(x+1)dx,

=(1/2)∫√(x+1)dx^2,

=(1/2)x^2√(x+1)-(1/2)∫x^2d√(x+1),

=(1/2)x^2√(x+1)-(1/4)∫x^2/√(x+1)dx,

=(1/2)x^2√(x+1)-(1/4)∫[x(x+1)-(x+1)+1]/√(x+1)dx,

=(1/2)x^2√(x+1)-(1/4)A+1/4∫√(x+1)dx-1/4∫dx/√(x+1),

(5/4)A=(1/2)x^2√(x+1)+1/4∫√(x+1)dx-1/2∫dx/2√(x+1),

A=(2/5)x^2√(x+1)+1/5∫√(x+1)d(x+1)-2/5√(x+1),

A=(2/5)x^2√(x+1)+2/15(x+1)^(3/2)-2/5*√(x+1)+C

咨询记录 · 回答于2022-03-17

∫（x+1）²dx求此不定积分

A=∫x√(x+1)dx,=(1/2)∫√(x+1)dx^2,=(1/2)x^2√(x+1)-(1/2)∫x^2d√(x+1),=(1/2)x^2√(x+1)-(1/4)∫x^2/√(x+1)dx,=(1/2)x^2√(x+1)-(1/4)∫[x(x+1)-(x+1)+1]/√(x+1)dx,=(1/2)x^2√(x+1)-(1/4)A+1/4∫√(x+1)dx-1/4∫dx/√(x+1),(5/4)A=(1/2)x^2√(x+1)+1/4∫√(x+1)dx-1/2∫dx/2√(x+1),A=(2/5)x^2√(x+1)+1/5∫√(x+1)d(x+1)-2/5√(x+1),A=(2/5)x^2√(x+1)+2/15(x+1)^(3/2)-2/5*√(x+1)+C

是∫（x+1）²dx

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