∫(log(x+1)-log x)/(x(x+1))dx is equal to:

$\int\frac{\log(x+1)-\log x}{x(x+1)}dx$

$\int(\log(x+1)-\log x).\frac{1}{x(x+1)}dx=\frac{-1}{2}[\log(\frac{x+1}{x})]^2+C$

$[\frac{d}{dx}[(x-1)-\log x]=\frac{1}{x+1}-\frac{1}{x}=-\frac{1}{(x+1)x}]$