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प्रश्न
\[\int\frac{\sec^2 x}{\tan x + 2} dx\]
बेरीज
उत्तर
\[\text{Let I} = \int\frac{\sec^2 x}{\ tanx + 2}dx\]
\[\text{Putting}\ \tan x = t\]
\[ \Rightarrow \sec^2 x = \frac{dt}{dx}\]
\[ \Rightarrow \sec^2 x dx = dt\]
\[ \therefore I = \int\frac{1}{t + 2}dt\]
\[ = \text{ln} \left| t + 2 \right| + C\]
\[ = \text{ln} \left| \ tan\ x + 2 \right| + C \left[ \because t = \tan x \right]\]
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