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Câu hỏi: Cho hàm số $f\left( x \right)=2{{x}^{2}}{{e}^{{{x}^{3}}+2}}+2x{{e}^{2x}}$, ta có $\int{f\left( x \right)dx}=m{{e}^{{{x}^{3}}+2}}+nx{{e}^{2x}}-p{{e}^{2x}}+C$. Giá trị của biểu thức $m+n+p$ bằng
A. $\dfrac{1}{3}$
B. 2
C. $\dfrac{13}{6}$
D. $\dfrac{7}{6}$
A. $\dfrac{1}{3}$
B. 2
C. $\dfrac{13}{6}$
D. $\dfrac{7}{6}$
Ta có: $I=\int{f\left( x \right)dx=\int{\left( 2{{x}^{2}}{{e}^{{{x}^{3}}+2}}+2x{{e}^{2x}} \right)dx=\int{\left( 2{{x}^{2}}{{e}^{{{x}^{3}}+2}} \right)dx+\int{\left( 2x{{e}^{2x}} \right)dx=J+K}}}}$
Tính $J=\int{\left( 2{{x}^{2}}{{e}^{{{x}^{3}}+2}} \right)dx}=\dfrac{2}{3}\int{\left( {{e}^{{{x}^{3}}+2}} \right)d\left( {{x}^{3}}+2 \right)}=\dfrac{2}{3}{{e}^{{{x}^{3}}+2}}+{{C}_{1}}$
Tính $K=\int{\left( 2x{{e}^{2x}} \right)dx}$
Đặt $\left\{ \begin{aligned}
& u=2x \\
& dx={{e}^{2x}}dx \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& du=2dx \\
& v=\dfrac{1}{2}{{e}^{2x}} \\
\end{aligned} \right.$
$\Rightarrow K=x.{{e}^{2x}}-\int{{{e}^{2x}}dx}=x{{e}^{2x}}-\dfrac{1}{2}{{e}^{2x}}+{{C}_{2}}$
Ta có: $I=J+K=\left( \dfrac{2}{3}{{e}^{{{x}^{3}}+2}}+{{C}_{1}} \right)+\left( x{{e}^{2x}}-\dfrac{1}{2}{{e}^{2x}}+{{C}_{2}} \right)=\dfrac{2}{3}{{e}^{{{x}^{3}}+2}}+x{{e}^{2x}}-\dfrac{1}{2}{{e}^{2x}}+C$
$\Rightarrow \left\{ \begin{aligned}
& m=\dfrac{2}{3} \\
& n=1 \\
& p=\dfrac{1}{2} \\
\end{aligned} \right.\Rightarrow m+n+p=\dfrac{13}{6}$
Tính $J=\int{\left( 2{{x}^{2}}{{e}^{{{x}^{3}}+2}} \right)dx}=\dfrac{2}{3}\int{\left( {{e}^{{{x}^{3}}+2}} \right)d\left( {{x}^{3}}+2 \right)}=\dfrac{2}{3}{{e}^{{{x}^{3}}+2}}+{{C}_{1}}$
Tính $K=\int{\left( 2x{{e}^{2x}} \right)dx}$
Đặt $\left\{ \begin{aligned}
& u=2x \\
& dx={{e}^{2x}}dx \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& du=2dx \\
& v=\dfrac{1}{2}{{e}^{2x}} \\
\end{aligned} \right.$
$\Rightarrow K=x.{{e}^{2x}}-\int{{{e}^{2x}}dx}=x{{e}^{2x}}-\dfrac{1}{2}{{e}^{2x}}+{{C}_{2}}$
Ta có: $I=J+K=\left( \dfrac{2}{3}{{e}^{{{x}^{3}}+2}}+{{C}_{1}} \right)+\left( x{{e}^{2x}}-\dfrac{1}{2}{{e}^{2x}}+{{C}_{2}} \right)=\dfrac{2}{3}{{e}^{{{x}^{3}}+2}}+x{{e}^{2x}}-\dfrac{1}{2}{{e}^{2x}}+C$
$\Rightarrow \left\{ \begin{aligned}
& m=\dfrac{2}{3} \\
& n=1 \\
& p=\dfrac{1}{2} \\
\end{aligned} \right.\Rightarrow m+n+p=\dfrac{13}{6}$
Đáp án C.