Integral from 0 to 1 of 1/(1+floor(1/x))

[@maxvangulik1988]

t=1/x
dx=-dt/t^2
I=int[1,♾️](t^-2/(1+floor(t)))dt
I=int[1,♾️](t^-2/ceil(t))dt
I=sum[n=1,♾️](int[n,n+1](t^-2/(n+1))dt)
I=sum[n=1,♾️](1/(n+1)•(-1/t)|[n,n+1])
I=-sum[n=1,♾️](1/(n+1)•(1/(n+1)-1/n))
I=1-pi^2/6+sum[n=1,♾️](1/n(n+1))
1/n(n+1)=1/n-1/(n+1)
I=1-pi^2/6+1
I=2-pi^2/6