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