If $\int x \log \left(1+\frac{1}{x}\right) d x = f(x) . \log _e(x+1)+g(x) \log _e x^2+L x+C$, then

$L=\frac{1}{2}$

Let

$I =\int x \log \left(1+\frac{1}{x}\right) d x$

$\Rightarrow I =\int x \log (1+x) d x-\int x \log x d x$

$\Rightarrow I=\frac{x^2}{2} \log (1+x)-\frac{1}{2} \int \frac{x^2}{x+1} d x-\left\{\frac{x^2}{2} \log _e x-\int \frac{1}{x} \times \frac{x^2}{2} d x\right\}$

$\Rightarrow I=\frac{x^2}{2} \log _e(1+x)-\frac{1}{2} \int x-1+\frac{1}{x+1} d x-\frac{x^2}{2} \log _e x+\frac{x^2}{4}+C$

$\Rightarrow I=\frac{x^2}{2} \log _e(1+x)-\frac{1}{2}\left(\frac{x^2}{2}-x\right)-\frac{1}{2} \log _e(x+1) -\frac{x^2}{2} \log _e x+\frac{x^2}{4}+C$

$\Rightarrow I=\left(\frac{x^2-1}{2}\right) \log _e(1+x)-\frac{x^2}{2} \log _e x+\frac{x}{2}+C$

Hence, $f(x)=\frac{x^2-1}{2}, g(x)=-\frac{x^2}{2}$ and $L=\frac{1}{2}$