Question

\[\int\frac{\log\left( 1 + \frac{1}{x} \right)}{x \left( 1 + x \right)} dx\]

Answer 1

\[\int\frac{\log \left( 1 + \frac{1}{x} \right)}{x\left( 1 + x \right)}dx\]

\[Let, \log \left( 1 + \frac{1}{x} \right) = t\]

\[ \Rightarrow \frac{1}{1 + \frac{1}{x}} \times \frac{- 1}{x^2} = \frac{dt}{dx}\]

\[ \Rightarrow \left( \frac{x}{x + 1} \right) \times \frac{- 1}{x^2} = \frac{dt}{dx}\]

\[ \Rightarrow \frac{- dx}{x\left( x + 1 \right)} = dt\]

\[ \Rightarrow \frac{dx}{x\left( x + 1 \right)} = - dt\]

\[Now, \int\frac{\log \left( 1 + \frac{1}{x} \right)}{x\left( 1 + x \right)}dx\]

= ∫ t   . (-dt)

\[ = \frac{- t^2}{2} + C\]

\[ = - \frac{1}{2} \left\{ \log\left( 1 + \frac{1}{x} \right) \right\}^2 + C\]