I=∫sinxdx/(3sinx+4cosx)=(1/5)∫sinxdx/sin[x+arcsin(4/5)]
=(1/5)∫sin[u-arcsin(4/5)]du/sinu
=(1/5)∫[(3/5)sinu-(4/5)cosu]du/sinu
=(1/25)∫(3-4cotu)du
=(1/25)(3u-4ln|sinu|)+C1
=(1/25){3[x+arcsin(4/5)]-4ln|sin[x+arcsin(4/5)]|)+C1
=(1/25)(3x-4ln|3sinx+4cosx|)+C.
答：当 c≤0 时，极限 lim[(x^3+7x^2-1)^c-x] 不存在。 当 c>0 时，极限 lim[(x^3+7x^2-1)^c-x] =lim[x^(3...详情>>
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