f(x)=xˆ2-lnxˆ2的单调区间 定积分ln(2x+1)dx (1,2e)

问题描述:

f(x)=xˆ2-lnxˆ2的单调区间 定积分ln(2x+1)dx (1,2e)

∫ln(2x+1)dx
=1/2∫ln(2x+1)d(2x+1)
=(2x+1)ln(2x+1)/2-1/2∫(2x+1)dln(2x+1)
=(2x+1)ln(2x+1)/2-1/2∫(2x+1)*1/(2x+1)*(2x+1)'dx
=(2x+1)ln(2x+1)/2-1/2∫2dx
=(2x+1)ln(2x+1)/2-x+C
=(4e+1)ln(4e+1)/2-3/2*ln3