Kết quả nguyên hàm I = ( 4x - 1).ln ^3( 2x)dx là: A. ( 2x^2 - x)ln ^3( 2x) - ( 3x^2 - 3x)ln ^2( 2x) - ( 3x^2 - 6x)ln ( 2x) + 3x^2/2 + 6x + C B. ( 2x^2 - x)ln ^3( 2x ) - ( 3x^2 - 3x)ln ^2( 2

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Kết quả nguyên hàm \[I = \int {\left( {4x - 1} \right).{{\ln }^3}\left( {2x} \right)dx} \] là:

\[\left( {2{x^2} - x} \right){\ln ^3}\left( {2x} \right) - \left( {3{x^2} - 3x} \right){\ln ^2}\left( {2x} \right) - \left( {3{x^2} - 6x} \right)\ln \left( {2x} \right) + \frac{{3{x^2}}}{2} + 6x + C\]

\[\left( {2{x^2} - x} \right){\ln ^3}\left( {2x} \right) - \left( {3{x^2} - 3x} \right){\ln ^2}\left( {2x} \right) + \left( {3{x^2} - 6x} \right)\ln \left( {2x} \right) - \frac{{3{x^2}}}{2} + 6x + C\]

\[\left( {2{x^2} - x} \right){\ln ^3}\left( {2x} \right) + \left( {3{x^2} - 3x} \right){\ln ^2}\left( {2x} \right) + \left( {3{x^2} - 6x} \right)\ln \left( {2x} \right) - \frac{{3{x^2}}}{2} + 6x + C\]

\[\left( {2{x^2} - x} \right){\ln ^3}\left( {2x} \right) + \left( {3{x^2} - 3x} \right){\ln ^2}\left( {2x} \right) + \left( {3{x^2} - 6x} \right)\ln \left( {2x} \right) - \frac{{3{x^2}}}{2} - 6x + C\]

Giải thích

Hướng dẫn giải

Media VietJack

Vậy \[I = \left( {2{x^2} - x} \right){\ln ^3}\left( {2x} \right) - \left( {3{x^2} - 3x} \right){\ln ^2}\left( {2x} \right) + \left( {3{x^2} - 6x} \right)\ln \left( {2x} \right) - \frac{{3{x^2}}}{2} + 6x + C\]

Chọn B.