[(1-u^2)/(u+u^3)]du=(1/x)dx求原函数怎么求啊?
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[(1-u^2)/(u+u^3)]du=(1/x)dx求原函数怎么求啊?
[(1-u^2)/(u+u^3)]du=(1/x)dx求原函数怎么求啊?
[(1-u^2)/(u+u^3)]du=(1/x)dx求原函数怎么求啊?
亲,见图
积分号(u^2-3)/(u^3-u) du=?
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x(x+1)du/dx=u^2;u(1)=1 求u(x)=?
[(1-u^2)/(u+u^3)]du=(1/x)dx求原函数怎么求啊?
-∫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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=2∫[u²/(1+u)]du=2∫[(u-1)+1/(u+1)]du 这一步是怎么求出来的.
{u+U=10,3u-2U=5