Use Integration by Parts
$\large\int \ln(x) dx$
set
$u = \ln(x),$ $dv = dx$
then we find
$du = \left(\frac{1}{x}\right) dx,$ $v = x$
substitute
$\large\int \ln(x) dx =\large\int u\; dv$
and use integration by parts
$= uv - \large\int v \;du$
substitute $u=\ln(x), v=x,$ and $du=\left(\frac{1}{x}\right)dx$
$= \ln(x)\; {x}- \large\int x \left(\frac{1}{x}\right) dx$
$= \ln(x) x -\large\int dx$
$= \ln(x) x - x + C$
$= x \ln(x) - x + C.$
Now Put Limits
$[\ln(1)-1+C]-[0-0+C]= -1$
Note-
$\lim [x\ln x] = 0.$
$x=0$
Correct Answer: $B$