If $\int x^{x}(1+\log x) d x=k x^{x}+c$, then $k=$
If $\int x^{x}(1+\log x) d x=k x^{x}+c$, then $k=$
$\log _{e} e$
$\log _{e}\left(\frac{1}{e^{2}}\right)$
$\log _{e}\left(e^{2}\right)$
$\log _{e}\left(\frac{1}{e}\right)$
Solution
Let $I=\int x^{x}(1+\log x) d x$
Put $x^{x}=t \Rightarrow e^{x \log x}=t \Rightarrow e^{x \log x}(1+\log x) d x=d t$ $\therefore x^{x}(1+\log x) d x=d t$
Thus, $I=\int d t=t+c$
$I=x^{x}+c$
Comparing with given data, we write $k=1=\log _{e} e$