Question

$\int x^x(1+\log x)dx$ is equal to

• A. $x^x+C$
• B. $x^{2x}+C$
• C. $x^x\log x+C$
• D. $1/2(1+\log x{)}^2+C$

Answer 1

Answer: (A)

$I=\int x^x(1+\log x)dx$
Put $x^x=t$, then $x^x(1+logx)dx=dt$
$\therefore I=\int dt\Rightarrow I=t+C\Rightarrow I=x^x+C$.