Tìm nguyên hàm \[I = \int {\frac{x}{{\sqrt {2x + 1} }}dx} .\]

Đáp án A

Đặt \(\sqrt {2x + 1} = t \Rightarrow x = \frac{{{t^2} - 1}}{2} \Rightarrow I = \int {\frac{{\frac{{{t^2} - 1}}{2}}}{t}d\left( {\frac{{{t^2} - 1}}{2}} \right) = \int {\frac{{{t^2} - 1}}.tdt} = \frac{1}{2}\int {\left( {{t^2} - 1} \right)dt} } \)

\( = \frac{1}{2}\left( {\frac{{{t^3}}}{3} - t} \right) + C = \frac{{{t^3}}}{6} - \frac{t}{2} + C = \frac{1}{6}\sqrt {{{\left( {2x + 1} \right)}^3}} - \frac{1}{2}\sqrt {2x + 1} + C.\)