求定积分ln(1+t)dt,上限e^x,下限-1的导数是多少,

∫ln(x+√(1+x^2))dx=xln(x+√(1+x^2))-∫xdln(x+√(1+x^2)
=xln(x+√(1+x^2)-√(1+x^2)+C
∫[0,1]ln(x+√(1+x^2)dx=ln(1+√2)-√2+1

用的是定积分的定义.
(ln(n)+ln(n+1)+...+ln(2n-1)-n·ln(n))/n
= (ln(1)+ln(1+1/n)+...+ln(1+(n-1)/n))/n
= ln(1)·1/n+ln(1+1/n)·1/n+...+ln(1+(n-1)/n)·1/n
就是ln(1+x)在[0,1]分划0 < 1/n < 2/n

∫(0->1/e) ln(1+x) dx
= [xln(1+x)](0->1/e) - ∫(0->1/e) [x/(1+x)] dx
=(1/e)[ ln(e+1) - 1] - ∫(0->1/e) dx + ∫(0->1/e) dx/(1+x)
=(1/e)[ ln(e+1) - 1] - [x](0->1/e) + [ln(1+x)](0->1/e)
=(1/e)[ ln(e+1) - 1] - 1/e + (ln(e+1) - 1)
= {[ ln(e+1) - 1] ( e+1) - 1}/e

∫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