A "proof" for $0=1$ by integrating $\int \frac{dx}{x\ln x}$ by parts 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. 
 A: 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.
