Answer
$a)$ $\int$$\ln(x)$$dx$ = $x\ln(x)-x+c$
$b$ $\frac{1}{e-1}$
Work Step by Step
$a)$
$\int$$\ln(x)$$dx$ = $x\ln(x)-x+c$
take diff both side
$\frac{d}{dx}$($\int$$\ln(x)$)$dx$ = $\frac{d}{dx}$($x\ln(x)-x+c$)
$\ln(x)$ = $x\frac{d}{dx}(\ln(x)+\ln(x)\frac{d}{dx}(x)-1$
$\ln(x)$ = $1+\ln(x)-1$
$\ln(x)$ = $\ln(x)$
$b)$
$avg$ = $\frac{1}{e-1}$$\int_{{\,1}}^{{\,e}}$$\ln(x)$ $dx$
$avg$ = $\frac{1}{e-1}$ ($x\ln(x)-x$)$|_{{\,1}}^{{\,e}}$
$avg$ = $\frac{1}{e-1}$ $[(e\ln(e)-e)-(\ln(1)-1)]$
$avg$ = $\frac{1}{e-1}$
