Question

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?

Answer 1

Integrating by parts with

`u=\frac1x\implies\mathrm du=-(\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.