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प्रश्न
उत्तर
` Here, we are" considering "log x as log_e x . `
\[\text{Let I} = \int\frac{1}{x\left( 3 + \log x \right)}dx\]
\[\text{Putting }\log x = t\]
\[ \Rightarrow \frac{1}{x} = \frac{dt}{dx}\]
\[ \Rightarrow \frac{dx}{x} = dt\]
\[ \therefore I = \int\frac{dt}{3 + t}\]
\[ = \text{log }\left| 3 + t \right| + C\]
\[ = \text{log }\left| 3 + \text{log x }\right| + C \left[ \because t = \text{log x} \right]\]
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