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