∫ x^x(1+log x) dx is equal to :

Let $I=\int x^x(1+\ln x) dx$

Let $x^{x}=t \Rightarrow x \ln x=\ln t$

$\Rightarrow\left(x . \frac{1}{x}+\ln x . 1\right)=\frac{1}{t} \frac{d t}{d x}$

$\Rightarrow dx(1+\ln x) x^{x}=dt$

∴  $I=\int d t \Rightarrow I=t+c$

$I=x^{x}+c$

Hence (2) is the correct answer.