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प्रश्न
उत्तर
\[\int\frac{\log \left( 1 + \frac{1}{x} \right)}{x\left( 1 + x \right)}dx\]
\[Let, \log \left( 1 + \frac{1}{x} \right) = t\]
\[ \Rightarrow \frac{1}{1 + \frac{1}{x}} \times \frac{- 1}{x^2} = \frac{dt}{dx}\]
\[ \Rightarrow \left( \frac{x}{x + 1} \right) \times \frac{- 1}{x^2} = \frac{dt}{dx}\]
\[ \Rightarrow \frac{- dx}{x\left( x + 1 \right)} = dt\]
\[ \Rightarrow \frac{dx}{x\left( x + 1 \right)} = - dt\]
\[Now, \int\frac{\log \left( 1 + \frac{1}{x} \right)}{x\left( 1 + x \right)}dx\]
= ∫ t . (-dt)
\[ = \frac{- t^2}{2} + C\]
\[ = - \frac{1}{2} \left\{ \log\left( 1 + \frac{1}{x} \right) \right\}^2 + C\]
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