Question

For $x>0$ and arbitrary constant of integration $C,$ $\int x^x(\frac{1}{x}+(\ln x{)}^2+\ln x)dx$ is equal to

• A. $x^x\left(\right(\ln x{)}^2-\frac{1}{x})+C$
• B. $x^x(\ln x-x)+C$
• C. $\frac{x^x(\ln x{)}^2}{2}+C$
• D. $x^x\ln x+C$

Answer 1

The correct option is D $x^x\ln x+C$

$I=\int x^x(\frac{1}{x}+(\ln x{)}^2+\ln x)dx$
$=\int x^x(\frac{1}{x}+(1+\ln x\left)\ln x\right)dx$
$=\int (x^x\frac{1}{x}+x^x(1+\ln x\left)\ln x\right)dx$
Recall $\int x^{x}dx=x^x(1+\ln x)$ and $\int u^{\prime }v+u{v}^{\prime }=uv$
$\therefore I=x^x\ln x+C$