Let's consider the indefinite integral $$\int \frac{dx}{x\ln x}.$$ We will compute it by integrating by parts: $$\int \frac{dx}{x\ln x}=\int (\ln x)'\frac{dx}{\ln x}=1+\int \frac{dx}{x\ln x}.$$ Hence, $0=1$. The question is : is there anything wrong in these computations?
Note: I came across this example while reading a book on counterexamples in real analysis. I thought that the people here would find this funny, especially because the error is not so obvious to everyone.
The issue here is what an indefinite integral means. An indefinite integral $\int f(x)dx$ is not a single function; rather, it's best thought of as a set of functions, namely, the set $$\{F: {d\over dx}F=f\}.$$ Elements of this set are called antiderivatives, or primitives, of $f$.
This set will always contain more than one element, since adding a constant won't affect the derivative. Notation like $$\int xdx={x^2\over 2}+C$$ is shorthand for "$\int xdx$ is the set of functions of the form $x\mapsto {x^2\over 2}+C$ for some $C\in\mathbb{R}$."
So no, your observation does not imply that zero equals one.
Note that it's not always the case that two antiderivatives must always differ by a constant: if the domain of the function is "not connected," we have more options. E.g. consider the function $f:x\mapsto {1\over x}$, defined on $\mathbb{R}-\{0\}$, has two "pieces" which can behave very differently: the function $F$ sending $x<0$ to $\ln(\vert x\vert) + 42$ and $x>0$ to $\ln( x) -17$ is an antiderivative of $f$, even though it doesn't differ from the "usual" antiderivative $x\mapsto\ln(\vert x\vert)$ by a constant.
