Question

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

Answer 1

$$\begin{gathered} \int \frac{\log \left(1+{\displaystyle \frac{1}{x}}\right)}{x\left(1+x\right)}dx \\ \mathrm{Let},\log \left(1+\frac{1}{x}\right)=t \\ \Rightarrow \frac{1}{1+{\displaystyle \frac{1}{x}}}\times \frac{-1}{x^2}=\frac{dt}{dx} \\ \Rightarrow \left(\frac{x}{x+1}\right)\times \frac{-1}{x^2}=\frac{dt}{dx} \\ \Rightarrow \frac{-dx}{x\left(x+1\right)}=dt \\ \Rightarrow \frac{dx}{x\left(x+1\right)}=-dt \\ \mathrm{Now},\int \frac{\log \left(1+{\displaystyle \frac{1}{x}}\right)}{x\left(1+x\right)}dx \\ =\int t·\left(-dt\right) \\ =\frac{-t^2}{2}+\mathrm{C} \\ =-\frac{1}{2}{\left\{\log \left(1+\frac{1}{\mathrm{x}}\right)\right\}}^2+\mathrm{C} \end{gathered}$$