Đáp án:

• Đặt \( t = \sqrt{2x + 1} \Rightarrow x = \frac{1}{2}(t^2 - 1) \Rightarrow dx = t \, dt \)

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

\( = \int \left( 2t^2 - 4t + 5 - \frac{10}{t + 2} \right) dt \)

\( = \frac{2t^3}{3} - 2t^2 + 5t - 10 \ln|t + 2| + C \)