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प्रश्न
\[\int\frac{\sqrt{16 + \left( \log x \right)^2}}{x} \text{ dx}\]
बेरीज
उत्तर
\[\text{ Let I }= \int\sqrt{\frac{16 + \left( \log x \right)^2}{x}}\text{ dx}\]
\[\text{ Putting log x }= t\]
\[ \Rightarrow \frac{1}{x} \text{ dx}= dt\]
\[ \therefore I = \int\sqrt{16 + t^2}dt\]
\[ = \int\sqrt{4^2 + t^2}dt\]
\[ = \frac{t}{2} \sqrt{4^2 + t^2} + \frac{4^2}{2} \text{ log} \left| t + \sqrt{4^2 + t^2} \right| + C\]
\[ = \frac{\log x}{2} \sqrt{16 + \left( \log x \right)^2} + 8 \text{ log }\left| \log x + \sqrt{16 + \left( \log x \right)^2} \right| + C\]
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