Question

\[\int\frac{\sec^2 x}{\tan x + 2} dx\]

Answer 1

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