An integral whose integrand blows up

Antiderivative. $(x\ln x-x)'=\ln x+1-1=\ln x$.

Computation. For $0<\varepsilon<1$,
$$\int_\varepsilon^1\ln x\,dx=\bigl[x\ln x-x\bigr]_\varepsilon^1=-1-(\varepsilon\ln\varepsilon-\varepsilon).$$
As $\varepsilon\to0^+$, $\varepsilon\ln\varepsilon\to0$ and $\varepsilon\to0$, so the limit is $-1$.

Answer. The integral converges, and $\displaystyle\int_0^1\ln x\,dx=-1$.

Remark. The integrand tends to $-\infty$ at $0$, yet the area is finite: $\ln x$ grows slower than any $x^{-p}$ with $p>0$, so the comparison with $\int_0^1x^{-1/2}dx<\infty$ already shows convergence.