Integrate : $$I=\int \frac{dx}{x\ln x}$$
I know this can be solved easily via substitution ($t=\ln x$) to obtain $I=\ln {\ln x} +C$, where $C$ is the constant of integration. But if we try to apply integration by parts, choosing $\frac{1}{\ln x}$ as $u(x)$ and $\frac{1}{x}$ as $v'(x)$, something goes wrong: $$I=\frac{1}{\ln x}\cdot\ln x - \int \left(\frac{-1}{(\ln x)^2}\cdot\frac{1}{x}\cdot\ln x\right)dx$$ This reduces to $I=1+I$ or $1=0$. Why does this happen?
You've shown $\int\frac{dx}{x\ln x}=1+\int\frac{dx}{x\ln x}$, which is no contradiction because $\int f^\prime(x)dx$ is the set of functions of the form $f(x)+C$, so it's analogous to $\Bbb R=1+\Bbb R$.
When we pass from indefinite integrals to definite ones, the rule $\int uv^\prime dx=uv-\int u^\prime vdx$ becomes $\int_a^b uv^\prime dx=[uv]_a^b-\int_a^b u^\prime vdx$, which in this case simplifies to $\int_a^b\frac{dx}{x\ln x}=0+\int_a^b\frac{dx}{x\ln x}$ because $[1]_a^b=0$.
