計(jì)算[0,ln2]∫√(e^x-1)dx
令√(e^x-1)=u,則e^x-1=u²,e^x=u²+1,e^xdx=2udu,dx=[2u/(u²+1)]du,
x=0時(shí)u=0,x=ln2時(shí)u=e^(ln2)-1=2-1=1
故原式=[0,1]2∫[u²/(u²+1)]du=[0,1]2∫[1-1/(u²+1)]du=2(u-arctanu)︱[0,1]
=2[1-π/4]=2-π/2.