Question

求积分∫ln{1+[(1+x)/x]^1/2}dx (x>0)

Answer 1

简单计算一下即可，答首肢芹历案如者首世图所示

Answer 2

分部积分，
原式=xln{1+[(1+x)/宴枝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)
带入化笑巧简碰祥键得到∫(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))]