Mathematics: Using integration by parts, ∫1/x dx = ∫1/x.1 dx = ∫1/x.x-∫(-1/x^2)x dx = ∫1+∫1/x dx. subtracting ∫1/x dx from both sideswe conclude that 0=1. what is wrong with this logic?

Using integration by parts, ∫1/x dx = ∫1/x.1 dx =

isalita

15.01.2022

Integrating by parts with

$u=\dfrac1x\implies\mathrm du=-\dfrac{\mathrm dx}{x^2}$

$\mathrm dv=\mathrm dx\implies v=x$

gives

$\displaystyle\int\frac{\mathrm dx}x=\frac xx+\int\frac x{x^2}\,\mathrm dx$

$\displaystyle\int\frac{\mathrm dx}x=1+\int\frac{\mathrm dx}x$

But the two integrals don't cancel, because there are infinitely many antiderivatives of $\frac1x$ that differ by a constant.

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