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प्रश्न
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उत्तर
\[\text{ Let I} = \int\frac{\log \left( \log x \right) dx}{x}\]
\[\text{ Putting log x = t}\]
\[ \Rightarrow \frac{1}{x} dx = dt\]
\[ \therefore I = \int 1_{II} \cdot \log _I t \cdot \text{ dt}\]
\[ = \log t\int1\text{ dt }- \int\left\{ \frac{d}{dt}\left( \log t \right)\int1 dt \right\}dt\]
\[ = \log t \cdot t - \int\frac{1}{t} \times t\text{ dt}\]
\[ = \log t \cdot t - \int dt\]
\[ = \log t \cdot t - t + C\]
\[ = t \left( \log t - 1 \right) + C\]
\[ = \log x \left( \text{ log} \left( \log x \right) - 1 \right) + C\]
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