ylny分之一的不定积分

∫1/(ylny)dy=∫1/lnydlny=lnlny+C

原式=∫1/(xlnx) dx=∫1/(lnx) dlnx=lnllnxl+C 绝对值很重要

直接凑微分计算:∫(ylny)^-1dy=∫[1/(ylny)]dy=∫(1/lny)dlny=ln|lny|+c.

∫1/tanydy=∫cosy/sinydy=∫1/sinydsiny=lnsiny+C

∫ lnydy = ylny-∫ ydlny = ylny-∫ y*(1/y)dy = ylny-∫ dy = ylny-y+C

y'sinx=ylny 分离变量,得dy/(ylny)=dx/sinx 两边同时积分,得ln/lny/=ln/cscx-cotx/+c1 整理得lny=c(cscx-cotx) 即y=e^c(cscx-cotx) (c为任意常数)

∫lnxdx=xlnx-∫x*1/xdx=x(lnx-1)+c 这里用到分步积分法

应该不用吧.

∫(1/sint)dt =∫[sint/(sint)^2]dt =-∫{1/[1-cost)(1+cost)]}d(cost) =-(1/2)∫[1/(1-cost)+1/(1+cost)]d(cost) =-(1/2)∫[1/(1-cost)]d(cost)-(1/2)∫[1/(1+cost)]d(cost) =(1/2)ln(1-cost)-(1/2)ln(1+cost)+C =(1/2)ln[(1-cost)/(1+cost)]+C =(1/2)ln[(1-cost)^2/(sint)^2]+C =ln|1/sint-cott|+C.