Question

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

Answer 1

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

Put $log(x+1)-logx=4(\frac{1}{x+1}-\frac{1}{x})dx=dt\frac{-1}{x(x+1)}dx-dt\frac{dx}{x(x+1)}=-dt$

$\int -tdt=\frac{-t^2}{2}+c=\frac{-{[log(x+1)-logx]}^2}{2}+c$