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Notice that
$$\left.\int_1^xt^ndt=\frac{t^{n+1}}{n+1}\right|_1^x=\frac{x^{n+1}-1}{n+1}$$
but at $n=-1$, we get division by $0$, and clearly this is bad. However, following this line of logic, we can take the limit as $n\to-1$ to get
$$\int_1^x\frac1tdt\stackrel?=\lim_{n\to-1}\frac{x^{n+1}-1}{...