Answer
$ \displaystyle \frac{2x+1}{x\ln 3(x+1)}$
Work Step by Step
Apply $\displaystyle \frac{d}{dx}[\log_{b}u]=\frac{\mathrm{l}}{u\ln b}\frac{du}{dx} \quad $(see p. 836)
$u(x)=x^{2}+x,$
$\displaystyle \frac{du}{dx}=2x+1.$
$\displaystyle \frac{d}{dx}[\log_{3}(x^{2}+x)]=\frac{\mathrm{l}}{(x^{2}+x)\ln 3}\cdot(2x+1)$
$=\displaystyle \frac{2x+1}{(x^{2}+x)\ln 3}$
$=\displaystyle \frac{2x+1}{x\ln 3(x+1)}$
