Tính I = nguyên hàm e^2x cos 3 x d x ta được:

15/20

Tính \[I = \smallint {e^{2x}}\cos 3xdx\] ta được:

\[\frac{{{e^{2x}}}}{{13}}\left( {2\sin 3x + 3\cos 3x} \right) + C\]

\[\frac{{{e^{2x}}}}{{13}}\left( {3\sin 3x - 2\cos 3x} \right) + C\]

\[\frac{{{e^{2x}}}}{{13}}\left( {2\sin 3x - 3\cos 3x} \right) + C\]

\[\frac{{{e^{2x}}}}{{13}}\left( {3\sin 3x + 2\cos 3x} \right) + C\]

Giải thích

Đặt

\(\left\{ {\begin{array}{*{20}{c}}{u = {e^{2x}}}\\{dv = cos3xdx}\end{array}} \right. \Rightarrow \left\{ {\begin{array}{*{20}{c}}{du = 2{e^{2x}}dx}\\{v = \frac{{sin3x}}{3}}\end{array}} \right. \Rightarrow I = \frac{1}{3}{e^{2x}}sin3x - \frac{2}{3}\smallint {e^{2x}}sin3xdx + {C_1}.\)

Xét nguyên hàm\[\smallint {e^{2x}}\sin 3xdx\] đặt

\(\left\{ {\begin{array}{*{20}{c}}{a = {e^{2x}}}\\{db = sin3xdx}\end{array}} \right. \Rightarrow \left\{ {\begin{array}{*{20}{c}}{da = 2{e^{2x}}}\\{b = - \frac{{cos3x}}{3}}\end{array}} \right.\)

\[ \Rightarrow \smallint {e^{2x}}\sin 3xdx = - \frac{1}{3}{e^{2x}}\cos 3x + \frac{2}{3}\smallint {e^{2x}}\cos 3x + {C_1} = - \frac{1}{3}{e^{2x}}\cos 3x + \frac{2}{3}I + {C_2}\]

Do đó ta có

\[\begin{array}{*{20}{l}}{I = \frac{1}{3}{e^{2x}}\sin 3x - \frac{2}{3}\left( { - \frac{1}{3}{e^{2x}}\cos 3x + \frac{2}{3}I + {C_2}} \right) + {C_1}}\\{ \Leftrightarrow \frac{{13}}{9}I = \frac{1}{3}{e^{2x}}\sin 3x + \frac{2}{9}{e^{2x}}\cos 3x + C}\\{ \Leftrightarrow I = \frac{1}{{13}}{e^{2x}}\left( {3\sin 3x + 2\cos 3x} \right) + C.}\end{array}\]

Đáp án cần chọn là: D