作业帮∫(u/(1+u-u^2-u^3)) du,求不定积分
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∫(u/(1+u-u^2-u^3)) du,求不定积分
∫(u/(1+u-u^2-u^3)) du,求不定积分
∫(u/(1+u-u^2-u^3)) du,求不定积分
∫udu/[(1+u)-(u^2+u^3)]
=∫udu/[(1+u)^2(1-u)]
=(1/2)∫[(1+u)-(1-u)]du/[(1+u)^2(1-u)]
=(1/2)∫du/[(1+u)(1-u)] -(1/2)∫du/(1+u)^2
=(1/4)∫[(1+u)+(1-u)]du/[(1+u)(1-u)] +(1/2)*(1/(1+u))
=(1/4)ln[|1+u|/|1-u|] +(1/2)*(1/(1+u))+C
∫(u/(1+u-u^2-u^3)) du,求不定积分
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-∫ud[u/(1+u)]=-u^2/(1+u)+∫u/(1+u)du=-u^2/(1+u) + ∫du -∫1/(1+u)d(u+1) = -u^2/(1+u)+u-ln|u+1|+C求救!特别是第一步到第二步之间!
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