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प्रश्न
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उत्तर
\[Let\ 1 + \log\ x\ = t . Then, \frac{1}{x}\ dx\ = dt\]
\[When\ x = 1, t = 1\ and\ x\ = 2, t = \left( 1 + \log 2 \right)\]
\[ \therefore I = \int_1^2 \frac{1}{x \left( 1 + \log x \right)^2} d x\]
\[ \Rightarrow I = \int_1^\left( 1 + \log 2 \right) \frac{1}{t^2} dt\]
\[ \Rightarrow I = \left[ \frac{- 1}{t} \right]_1^\left( 1 + \log 2 \right) \]
\[ \Rightarrow I = - \frac{1}{\left( 1 + \log 2 \right)} + 1\]
\[ \Rightarrow I = \frac{\log 2}{\log\ 2 + \log e}\]
\[ \Rightarrow I = \frac{\log 2}{\log\ 2e}\]
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