Đáp án:

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

• \(I = \int \frac{t^4}{1 - t^2} \, dt = \int \frac{t^4 - 1 + 1}{1 - t^2} \, dt = \int \left( -t^2 - 1 + \frac{1}{1 - t^2} \right) \, dt\)

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

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

\(= -\frac{t^3}{3} - t + \frac{1}{2} \ln \left| \frac{1 + t}{1 - t} \right| + C\)