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

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

人气：463 ℃　时间：2020-06-02 00:11:03

解答

∫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