Using integration by parts, let $$f(x)=\ln x\implies f'(x)=\frac1x$$ and $$g'(x)=1\implies g(x)=x$$ Then $$\int_0^1\ln x\,dx=[x\ln x]_0^1-\int_0^11\,dx=-1$$ Since each of the four quarters are identical, we have that the total area is $$I=4|-1|=\color{red}{4}$$
$$\int_0^1\ln x\mathrm{d}x=x\ln x-x\bigg|_0^1=-1$$ $$Area=-4\int_0^1\ln x\mathrm{d}x=4$$