Question

Evaluate :

$\int x^x(1+\log x)dx$

Answer 1

Let $x^x=t$. Then,

$d\left(x^x\right)=dt$

$d\left(e^{x\log x}\right)=dt$

$e^{x\log x}(\log x+1)dx=dt$

$x^x(1+\log x)dx=dt$

Therefore, $I=\int x^x(1+\log x)dx$

$I=\int dt=t+C=x^x+C$