Giả sử $\Large \int{{{e}^{2x}}(2{{x}^{3}}+5{{x}^{2}}-2x+4)dx=(a{{x}^{3

Giả sử $\Large \int{{{e}^{2x}}(2{{x}^{3}}+5{{x}^{2}}-2x+4)dx=(a{{x}^{3}}+b{{x}^{2}}+cx+d){{e}^{2x}}+C}$ . Khi đó $\Large a+b+c+d$ bằng

• A.

  $\Large -2$

• B.

  $\Large 3$

• C.

  $\Large 2$

• D.

  $\Large 5$

Ta có $\Large \int{{{e}^{2x}}(2{{x}^{3}}+5{{x}^{2}}-2x+4)dx=(a{{x}^{3}}+b{{x}^{2}}+cx+d){{e}^{2x}}+C}$ nên :

$\large ((a{{x}^{3}}+b{{x}^{2}}+cx+d){{e}^{2x}}+C{)}'$$\large =(3a{{x}^{2}}+2bx+c){{e}^{2x}}+2{{e}^{2x}}(a{{x}^{3}}+b{{x}^{2}}+cx+d)$

$\Large =(2a{{x}^{3}}+(3a+2b){{x}^{2}}+(2b+2c)x+c+2d){{e}^{2x}}$

$\Large =(2{{x}^{3}}+5{{x}^{2}}-2x+4){{e}^{2x}}$

Do đó: $\Large \left\{ \begin{align}  & 2a=2 \\  & 3a+2b=5 \\  & 2b+2c=-2 \\  & c+2d=4 \\ \end{align} \right.$ $\Large \Leftrightarrow \left\{ \begin{align}  & a=1 \\  & b=1 \\  & c=-2 \\  & d=3 \\ \end{align} \right.$

Vậy $\Large a+b+c+d=3$

Chọn đáp án B