This example is due to Vladimir Nikolin.

We are going to evaluate \int \frac{dx}{x} by partial integration:

\int udv = uv - \int vdu.

For this example, set u=\frac{1}{x}, v=x, then du=-\frac{1}{x^2}dx, dv=dx. So we obtain,

\int\frac{dx}{x} = x\cdot \frac{1}{x} - \int x\cdot \big (-\frac{1}{x^2}\big )dx.
\int\frac{dx}{x} = 1 + \int\frac{dx}{x}.
A = 1 + A.
0=1.