Đáp án:

• \(I = \int \frac{1}{\cos^2 x}\cdot \frac{1}{ (\tan^2 x - 3\tan x - 4)} \, dx \quad \text{đặt } t = \tan x\)

\(= \int \frac{1}{t^2 - 3t - 4} \, dt = \int \frac{1}{(t + 1)(t - 4)} \, dt\)

\(= \frac{1}{5} \int \left( \frac{1}{t - 4} - \frac{1}{t + 1} \right) \, dt = \frac{1}{5} \ln \left| \frac{t - 4}{t + 1} \right| + C\)