$$\int_0^{3}\frac{dx}{x-1}$$
I was looking at this problem, and have not seen dx ever go on top, how might I approach this problem?
2026-04-24 22:26:39.1777069599
Definite integral with dx on top
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Just as $$3\cdot\frac{1}{4}=\frac{3}{4}$$ $$\int_0^3 \frac{1}{x-1}\,dx=\int_0^3 \frac{dx}{x-1}$$ I will also note that this integral doesn't even converge in the usual sense, unless you look at its Cauchy Principal Value.