Question
$\text{Evaluate} \int \frac{\sec^{2} x}{3 + \tan x} \text{dx} $
$\text{Let 3} + \tan x = t$
$\sec^{2} x dx = dt$
$\therefore \int \frac{ \sec^{2} x}{3+ \tan x} dx = \int\frac{dt}{t}$
$= \log |t|+ c$
$= \log|3 + \tan x| + c$
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