Question: The function $f(x)=x^3$ has the derivative $3x^2$ which is equal to $0$ at $x=0$ but the function $f(x)=x^3$ has neither maxima nor minima at that point. How is this possible?

Answer: 
Recall that $\ln x$ is the inverse of the function $e^x$. We have the following identity for the integrals of inverse functions:

$$\int f^{-1}(x)\,dx=xf^{-1}(x)-F\circ f^{-1}(x)+C,$$

where $F$ is the antiderivative of $f$, not $f^{-1}$. Using this, we obtain the integral of $\ln x$ very easily:

$$\int\ln x\,dx=x\ln x-e^{\ln x}+C=x\ln x-x+C$$
Just calculate $$\left(x\ln{x}-x\right)'.$$