[(1-u^2)/(u+u^3)]du=(1/x)dx求原函数怎么求啊?

积分号(u^2-3)/(u^3-u) du=? ∫(u/(1+u-u^2-u^3)) du,求不定积分 1/(u+u^2)du求不定积分 ∫(u+2)/(u^2+3u)du积分 求不定积分 ∫(u-1)(u^2+u+1)du 求不定积分 ∫(u-1)(u^2+u+1)du 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! ∫2/(1-u^2+2u)du怎么做 du/[u(u-1)]=dx/x详解 du/dx=(x+u)^2求u的解 du/(u-u^2)=dx/x怎么解, 不定积分根号(u-1/u)du du/（u^2-u）积分 du/(u-u^2)怎么积分 =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