Chapter 8.2, Problem 50E

Using integration by parts,

$\begin{array}{rll}{\displaystyle \int \frac{1}{x}}dx& =& {\displaystyle \int \frac{1}{x}}\cdot 1dx\\ & =& \frac{1}{x}\cdot x-{\displaystyle \int \left(-\frac{1}{x^2}\right)}xdx\\ & =& 1+{\displaystyle \int \frac{1}{x}}dx\end{array}$

Subtracting ${\displaystyle \int \frac{1}{x}}dx$ from both sides we conclude that $0=1$.

What is wrong with this logic ? Source: Sam Northshield.