∫√(1+e x) dx在（ln3 ln8）定积分

问题补充：

∫√(1+e x) dx在（ln3,ln8）定积分

答案：

答：设√(1+e^x)=t>1,1+e^x=t^2,x=ln(t^2-1)

x=ln3,t=2

x=ln8,t=3

原式=(2→3) ∫td[ln(t^2-1)]

=(2→3) ∫ [t*2t/(t^2-1)]dt

=(2→3) 2∫ [(t^2-1+1)/(t^2-1)]dt

=(2→3) 2∫ [1+1/(t^2-1)]dt

=(2→3) 2*[t+(1/2)ln|(x-1)/(x+1)|]

=2*[3+(1/2)ln(1/2)]-2*[2+(1/2)ln(1/3)]

=6-ln2-4+ln3

=2+ln3-ln2

=2+ln(3/2)