Integrate the function $\frac{1}{x+x \log x}$.

(N/A) The given function can be rewritten as:
$\frac{1}{x+x \log x} = \frac{1}{x(1+\log x)}$
Let $1+\log x = t$.
Then,differentiating both sides with respect to $x$,we get:
$\frac{1}{x} dx = dt$
Substituting these into the integral:
$\int \frac{1}{x(1+\log x)} dx = \int \frac{1}{t} dt$
Integrating with respect to $t$:
$= \log |t| + C$
Substituting back $t = 1 + \log x$:
$= \log |1 + \log x| + C$
where $C$ is an arbitrary constant.