求积分∫ln{1+[(1+x)/x]^1/2}dx (x>0)

1个回答

  • 分部积分,

    原式=xln{1+[(1+x)/x]^1/2}-∫(-1/2)sqrt(x/(1+x))/x(1+sqrt((1+x)/x)dx

    考虑后面的部分,令u=sqrt((1+x)/x),x=1/(u^2-1)

    带入化简得到∫(1/2-1/2u)2udu/(1-u^2)^2=-∫du/(1+u)(1-u^2)=(-1/2)∫du/(1+u)^2-(1/4)∫du/(1-u)-(1/4)∫du/(1+u)=1/2(1+u)+(1/4)ln[(1-u)/(1+u)]

    原式=xln{1+[(1+x)/x]^1/2}+1/2(1+sqrt((1+x)/x))+(1/4)ln[(1-sqrt((1+x)/x))/(1+sqrt((1+x)/x))]