Evaluating a certain integral without the fundamental theorem

For logarithmic integrals a subdivision into geometric progression is often convenient. Set $r=\sqrt[n]{x}$ and consider the upper sum $$ \sum_{k=1}^n (r^k-r^{k-1})\ln(r^k)= \ln r\sum_{k=1}^n k(r^k-r^{k-1}) $$ It's easy to show, by induction, that $$ \sum_{k=1}^n k(r^k-r^{k-1})=nr^n-\sum_{k=0}^{n-1}r^k =nr^n-\frac{r^n-1}{r-1} $$ Putting back $r=x^{1/n}$, we get, for the upper sum, the expression $$ \left(x-\frac{x-1}{n(x^{1/n}-1)}\right)\ln x $$ Now, $$ \lim_{n\to\infty}n(x^{1/n}-1)=\lim_{t\to0^+}\frac{x^t-1}{t}=\ln x $$ so the limit of the upper sums is $$ \left(x-\frac{x-1}{\ln x}\right)\ln x=x\ln x-x+1 $$

Check similarly for the lower sums and see that this agrees with $$ \int_1^x\ln t\,dt=x\ln x-x+1 $$