If ${\displaystyle \int {\log }_{10}xdx}=K.x\log f\left(x\right)+c$  then $K,f\left(x\right)=$

${\displaystyle \int \frac{\log x}{\log 10}dx}$ $=\frac{1}{\log 10}\left[\left(\log x\right)x-{\displaystyle \int x\frac{1}{x}dx}\right]$

                      $=\frac{1}{\log 10}\left[x\log x-x\right]+c$

                      $=\left({\log }_{10}^e\right)x\left[\log x-\log e\right]+c$

        $K={\log }_{10}^e,f\left(x\right)=\left(\frac{x}{e}\right)$