$\int \frac{1}{\log \left(x^x\right)(\log x+1)}dx=$

The given integral is $\int \frac{1}{\log \left(x^x\right)(\log x+1)}dx$

Use the logarithmic property $\log \left(x^x\right)=x\log x$, the given integral can be written as 

$\int \frac{1}{\log \left(x^x\right)(\log x+1)}dx=\int \frac{1}{x\log x\left(\log \left(x+1\right)\right)}dx$

Suppose that $\log x=t$$\Rightarrow \frac{1}{x}dx=dt$

Hence, the given intergal becomes $\int \frac{1}{t\left(1+t\right)}dt$

Use partial fractions to find this integral

hence, 

   $\begin{array}{rcl}\int \frac{1}{t\left(1+t\right)}dt& =& \int \left(\frac{1}{t}-\frac{1}{1+t}\right)dt\\ & =& \ln t-\ln \left(1+t\right)\\ & =& \ln \left(\log \left|x\right|\right)-\ln \left(1+\log x\right)+C\\ & =& \log \left(\frac{\log x}{1+\log x}\right)+C\end{array}$

Therefore, $\int \frac{1}{\log \left(x^x\right)(\log x+1)}dx=\boxed{\log \left|\frac{\log x}{1+\log x}\right|+C}$