Integrating by parts with
[tex]u=\dfrac1x\implies\mathrm du=-\dfrac{\mathrm dx}{x^2}[/tex]
[tex]\mathrm dv=\mathrm dx\implies v=x[/tex]
gives
[tex]\displaystyle\int\frac{\mathrm dx}x=\frac xx+\int\frac x{x^2}\,\mathrm dx[/tex]
[tex]\displaystyle\int\frac{\mathrm dx}x=1+\int\frac{\mathrm dx}x[/tex]
But the two integrals don't cancel, because there are infinitely many antiderivatives of [tex]\frac1x[/tex] that differ by a constant.
