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प्रश्न
\[\int\left( \frac{1}{\log x} - \frac{1}{\left( \log x \right)^2} \right) dx\]
योग
उत्तर
\[\text{ Let I }= \int\left( \frac{1}{\log x} - \frac{1}{\left( \log x \right)^2} \right)dx\]
\[\text{ Put log x }= t\]
\[ \Rightarrow x = e^t \]
\[ \Rightarrow dx = e^t dt\]
\[ \therefore I = \int e^t \left( \frac{1}{t} - \frac{1}{t^2} \right)dt\]
\[\text{ Here}, f(t) = \frac{1}{t}\]
\[ \Rightarrow f'(t) = \frac{- 1}{t^2}\]
\[\text{ let e} ^t \times \frac{1}{t} = p\]
\[\text{ Diff both sides w . r . t t}\]
\[\left( e^t \times \frac{1}{t} + e^t \times \frac{- 1}{t^2} \right)dt = dp\]
\[ \therefore I = \int dp\]
\[ = p + C\]
\[ = \frac{e^t}{t} + C\]
\[ = \frac{x}{\log x} + C\]
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